A triangle has corners at #(5 ,2 )#, #(2 ,7 )#, and #(3 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(5  ,2  )#, #(2  ,7  )#, and #(3   ,5  )#.  How far is the triangle's centroid from the origin?

Sadie Allen · Accepted Answer

The triangle's centroid is #5.735# units from the origin.

Coordinates of centroid of a triangle whose vertices (corners) are #(x_1,y_1)#, #(x_2,y_2)# and #(x_3,y_3)# is given by #((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)# As corners of triangle are #(5,2)#, #(2,7)# and #(3,5)#, the centroid of given triangle is #((5+2+3)/3,(2+7+5)/3)# or #(10/3,14/3)#. And its distance from origin is #sqrt((10/3-0)^2+(14/3-0)^2)=sqrt(100/9+196/9)=sqrt(296/9)# = #1/3sqrt2xx2xx74=2/3xxsqrt74=2/3xx8.602=5.735#

Wyatt Anderson · Answer

undefined To find the centroid of a triangle, the coordinates of the centroid $ (x_c, y_c) $ can be calculated using the formula:

$$ x_c = \frac{x_1 + x_2 + x_3}{3} $$
$$ y_c = \frac{y_1 + y_2 + y_3}{3} $$

Given the coordinates of the triangle's vertices as $ (5, 2) $, $ (2, 7) $, and $ (3, 5) $, the centroid's coordinates can be found as follows:

$$ x_c = \frac{5 + 2 + 3}{3} = \frac{10}{3} $$
$$ y_c = \frac{2 + 7 + 5}{3} = \frac{14}{3} $$

Thus, the centroid of the triangle is located at $ \left(\frac{10}{3}, \frac{14}{3}\right) $.

To find the distance between the centroid and the origin, we can use the distance formula:

$$ d = \sqrt{(x_c - x_o)^2 + (y_c - y_o)^2} $$

where $ (x_o, y_o) $ represents the coordinates of the origin, which is $ (0, 0) $ in this case.

$$ d = \sqrt{\left(\frac{10}{3} - 0\right)^2 + \left(\frac{14}{3} - 0\right)^2} $$
$$ d = \sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{14}{3}\right)^2} $$
$$ d = \sqrt{\frac{100}{9} + \frac{196}{9}} $$
$$ d = \sqrt{\frac{296}{9}} $$
$$ d = \frac{\sqrt{296}}{3} $$

Therefore, the distance between the centroid of the triangle and the origin is $ \frac{\sqrt{296}}{3} $ units.