A triangle has corners at #(7 ,9 )#, #(1 ,1 )#, and #(3 ,8 )#. How far is the triangle's centroid from the origin?

Answer 1

Triangle's centroid is #7.032# units away from the origin.

Centroid of a triangle, whose corners are #(x_1,y_1)#, #(x_2,y_2)# and #(x_3,y_3)#, is given by #(1/3(x_1+x_2+x_3),1/3(y_1+y_2+y_3))#
Hence centroid of the triangle whose corners are #(7,9)#, #(1.1)# and #(3,8)# is
#(1/3(7+1+3),1/3(9+1+8))# or #(11/3,18/3)#
And its distance from origin #(0,0)# is
#sqrt((18/3-0)^2+(11/3-0)^2)=sqrt(324/9+121/9)#
= #sqrt(445/9)=1/3sqrt445=1/3xx21.095=7.032#
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Answer 2

To find the centroid of a triangle, you can average the coordinates of its vertices.

The coordinates of the centroid (G) can be calculated using the formula: [ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) ]

Given the vertices: [ A(7, 9), B(1, 1), C(3, 8) ]

[ G_x = \frac{7 + 1 + 3}{3} = \frac{11}{3} ] [ G_y = \frac{9 + 1 + 8}{3} = \frac{18}{3} = 6 ]

So, the centroid G is at ((\frac{11}{3}, 6)).

Now, to find the distance between the centroid and the origin, use the distance formula: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

[ d = \sqrt{(\frac{11}{3} - 0)^2 + (6 - 0)^2} ] [ d = \sqrt{\left(\frac{11}{3}\right)^2 + 6^2} ] [ d = \sqrt{\frac{121}{9} + 36} ] [ d = \sqrt{\frac{121 + 324}{9}} ] [ d = \sqrt{\frac{445}{9}} ] [ d \approx \sqrt{49.44} ] [ d \approx 7.04 ]

Therefore, the distance between the centroid and the origin is approximately (7.04) units.

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Answer 3

To find the centroid of a triangle, you average the coordinates of its vertices. The centroid of a triangle with vertices ( (x_1, y_1) ), ( (x_2, y_2) ), and ( (x_3, y_3) ) is given by the coordinates ( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) ).

Given the vertices of the triangle as ( (7, 9) ), ( (1, 1) ), and ( (3, 8) ), the centroid's coordinates are:

[ \left(\frac{7 + 1 + 3}{3}, \frac{9 + 1 + 8}{3}\right) = \left(\frac{11}{3}, \frac{18}{3}\right) = \left(\frac{11}{3}, 6\right) ]

Now, to find the distance from the centroid to the origin, which is at ( (0, 0) ), you can use the distance formula:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

where ( (x_1, y_1) ) is the coordinates of the centroid and ( (x_2, y_2) ) is the coordinates of the origin.

Plugging in the values, we get:

[ \text{Distance} = \sqrt{\left(0 - \frac{11}{3}\right)^2 + \left(0 - 6\right)^2} = \sqrt{\left(\frac{-11}{3}\right)^2 + (-6)^2} = \sqrt{\frac{121}{9} + 36} = \sqrt{\frac{121}{9} + \frac{324}{9}} = \sqrt{\frac{445}{9}} ]

So, the distance from the centroid to the origin is ( \sqrt{\frac{445}{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.

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