A triangle has corners at #(-3 ,-1 )#, #(7 ,3 )#, and #(1 ,2 )#. If the triangle is dilated by a factor of #1/3 # about point #(-3 ,4 ), how far will its centroid move?
Answer 1
Wyatt Anderson#A(-3,-1), B(7,3), C(1, 2), " about point " D (-3, 4), " dilation factor "1/3#
Centroid #G(x,y) = ((x_a + x_b + x_c) /3, (y_a + y_b + y_c)/3)#
#G(x,y) = ((-3 + 7+ 1)/3, (-1 + 3 + 2)/3) = (5/3, 4/3)#
#A'((x),(y)) = 1/3a - -2/3d = 1/3*((-3),(-1)) - -2/3*((-3),(4)) = ((-3),(7/3))#
#B'((x),(y)) = 1/3b - -2/3d = 1/3*((7),(3)) - -2/3*((-3),(4)) = ((1/3),(11/3))#
#C'((x),(y)) = 1/3c - -2/3d = 1/3*((1),(2)) - -2/3*((-3),(4)) = ((-5/3),(10/3))#
#"New Centroid " G'(x,y) = ((-3+ 1/3 - 5/3)/3,(7/3 + 11/3 + 10/3)/3) = (-7/9,28/9)#
#color(indigo)("Distance moved by centroid " #
#color(crimson)(vec(GG') = sqrt((5/3- -7/9)^2 + (4/3 -28/9)) ~~ 3.02255 " units"#
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Answer 2
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to findTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilationTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. TheTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroidTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid willTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + yTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will moveTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move byTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the sameTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factorTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilationTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation appliedTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied toTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the verticesTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let'sTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculateTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} =To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the originalTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. AverageTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of xTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinatesTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates:To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3},To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right)To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. AverageTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinatesTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{CentTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{CentroidTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} =To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3},To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right)To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
SoTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So,To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
AfterTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation byTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroidTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by aTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factorTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor ofTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the originalTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangleTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle isTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is atTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \rightTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $,To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right)To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, theTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroidTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves inTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now,To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the sameTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, letTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratioTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let'sTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. SoTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the verticesTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation byTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by aTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factorTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{NewTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New CentTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New CentroidTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} =To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \timesTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ aboutTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the pointTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3,To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4)To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right)To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. ForTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the firstTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New CentTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
\To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ xTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9},To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
NowTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now,To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, findTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3))To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find theTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distanceTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance betweenTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the newTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroidTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ yTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} =To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrtTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4)To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} -To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the secondTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertexTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7,To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
\To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 }To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3))To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrtTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ yTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} -To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4)To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} -To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
\To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ xTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 }To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3))To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \leftTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left(To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y'To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \rightTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{DistanceTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} =To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 -To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrtTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} +To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
NowTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now,To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, letTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} }To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let'sTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \textTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of theTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the newTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrtTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1.To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. AverageTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average ofTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of newTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new xTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinatesTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates:To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} }To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{DistanceTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \fracTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrtTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} +To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9}To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
ThereforeTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} =To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore,To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, theTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \fracTo find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, the centroidTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \frac{2To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, the centroid movesTo find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \frac{2}{To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, the centroid moves a distance of $ \frac{\sqrt{164}}{9} \To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \frac{2}{3To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, the centroid moves a distance of $ \frac{\sqrt{164}}{9} $To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \frac{2}{3}To find the centroid of a triangle, you average the coordinates of its vertices. Then, after dilation, the centroid moves in a consistent ratio with the dilation factor.
First, find the centroid of the original triangle:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$
$$ \text{Centroid} = \left( \frac{-3 + 7 + 1}{3}, \frac{-1 + 3 + 2}{3} \right) $$
$$ \text{Centroid} = \left( \frac{5}{3}, \frac{4}{3} \right) $$
After dilation by a factor of $ \frac{1}{3} $, the centroid moves in the same ratio. So the new centroid coordinates are:
$$ \text{New Centroid} = \left( \frac{1}{3} \times \frac{5}{3}, \frac{1}{3} \times \frac{4}{3} \right) $$
$$ \text{New Centroid} = \left( \frac{5}{9}, \frac{4}{9} \right) $$
Now, find the distance between the original centroid and the new centroid:
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{5}{3} \right)^2 + \left( \frac{4}{9} - \frac{4}{3} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( \frac{5}{9} - \frac{15}{9} \right)^2 + \left( \frac{4}{9} - \frac{12}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \left( -\frac{10}{9} \right)^2 + \left( -\frac{8}{9} \right)^2 } $$
$$ \text{Distance} = \sqrt{ \frac{100}{81} + \frac{64}{81} } $$
$$ \text{Distance} = \sqrt{ \frac{164}{81} } $$
$$ \text{Distance} = \frac{\sqrt{164}}{9} $$
Therefore, the centroid moves a distance of $ \frac{\sqrt{164}}{9} $ units.To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, after dilation, you repeat the process to find the new centroid. The centroid will move by the same factor as the dilation applied to the vertices.
Let's calculate the centroid of the original triangle:
1. Average of x-coordinates: $$ \frac{-3 + 7 + 1}{3} = \frac{5}{3} $$
2. Average of y-coordinates: $$ \frac{-1 + 3 + 2}{3} = \frac{4}{3} $$
So, the centroid of the original triangle is at $ \left( \frac{5}{3}, \frac{4}{3} \right) $.
Now, let's calculate the new coordinates of the vertices after dilation by a factor of $ \frac{1}{3} $ about the point $ (-3, 4) $:
1. For the first vertex (-3, -1):
$$ x' = -3 + \frac{1}{3}(-3 - (-3)) = -3 $$
$$ y' = 4 + \frac{1}{3}(-1 - 4) = \frac{11}{3} $$
2. For the second vertex (7, 3):
$$ x' = -3 + \frac{1}{3}(7 - (-3)) = \frac{8}{3} $$
$$ y' = 4 + \frac{1}{3}(3 - 4) = \frac{11}{3} $$
3. For the third vertex (1, 2):
$$ x' = -3 + \frac{1}{3}(1 - (-3)) = 1 $$
$$ y' = 4 + \frac{1}{3}(2 - 4) = \frac{10}{3} $$
Now, let's find the centroid of the new triangle:
1. Average of new x-coordinates: $$ \frac{-3 + \frac{8}{3} + 1}{3} = \frac{2}{3} $$
2. Average of new y-coordinates: $$ \frac{\frac{11}{3} + \frac{11}{3} + \frac{10}{3}}{3} = \frac{32}{9} $$
So, the centroid of the new triangle is at $ \left( \frac{2}{3}, \frac{32}{9} \right) $.
To find how far the centroid has moved, we calculate the distance between the original and new centroids:
$$ \text{Distance} = \sqrt{\left( \frac{2}{3} - \frac{5}{3} \right)^2 + \left( \frac{32}{9} - \frac{4}{3} \right)^2} $$
$$ \text{Distance} = \sqrt{\left( -\frac{3}{3} \right)^2 + \left( \frac{32}{9} - \frac{12}{9} \right)^2} $$
$$ \text{Distance} = \sqrt{\left( -1 \right)^2 + \left( \frac{20}{9} \right)^2} $$
$$ \text{Distance} = \sqrt{1 + \frac{400}{81}} $$
$$ \text{Distance} = \sqrt{\frac{481}{81}} $$
$$ \text{Distance} = \frac{\sqrt{481}}{9} $$
Therefore, the centroid moves a distance of $ \frac{\sqrt{481}}{9} $ units.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
Trending questions
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