A triangle has corners at #(4 ,1 )#, #(8 ,3 )#, and #(3 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(4  ,1  )#, #(8  ,3  )#, and #(3   ,5  )#.  How far is the triangle's centroid from the origin?

Jayden Jackson · Accepted Answer

Triangle’s centroid is 5.83 away from origin.

Centroid
#G (x) = (x1 + x2 + x3) / 3 = (4+8+3)/3 = 5#

#G(y) = (y1 + y2 + y3)/3 = (1+3+5)/3 = 3#

Coordinates of centroid G is #(5,3)#

Distance from origin #d = sqrt(x^2 + y2) = sqrt(5^2 + 3^2) ~~ color(red)(5.83)#

Noah Atkinson · Answer

undefined The centroid of a triangle is the point where the three medians intersect. To find the centroid, we calculate the average of the coordinates of the three vertices. Then, we find the distance between the centroid and the origin using the distance formula.

Given the vertices of the triangle are (4, 1), (8, 3), and (3, 5), the centroid can be found by averaging the x-coordinates and the y-coordinates separately.

The x-coordinate of the centroid, $ \bar{x} $, is the average of the x-coordinates of the vertices:
$$ \bar{x} = \frac{4 + 8 + 3}{3} = \frac{15}{3} = 5 $$

The y-coordinate of the centroid, $ \bar{y} $, is the average of the y-coordinates of the vertices:
$$ \bar{y} = \frac{1 + 3 + 5}{3} = \frac{9}{3} = 3 $$

So, the centroid of the triangle is at (5, 3).

Now, we calculate the distance between the centroid (5, 3) and the origin (0, 0) using the distance formula:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
$$ d = \sqrt{(5 - 0)^2 + (3 - 0)^2} $$
$$ d = \sqrt{5^2 + 3^2} $$
$$ d = \sqrt{25 + 9} $$
$$ d = \sqrt{34} $$

So, the distance between the centroid of the triangle and the origin is $ \sqrt{34} $.