A triangle has corners at #(7 ,9 )#, #(1 ,4 )#, and #(3 ,8 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(7  ,9  )#, #(1  ,4  )#, and #(3   ,8  )#.  How far is the triangle's centroid from the origin?

Sadie Allen · Accepted Answer

Triangle's centroid from the origin is #7.902# units away.

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)#, #(1.4)# and #(3,8)# is #(1/3(7+1+3),1/3(9+4+8))# or #(11/3,21/3)# And its distance from origin #(0,0)# is #sqrt((11/3-0)^2+(21/3-0)^2)=sqrt(121/9+441/9)# = #1/3sqrt562=1/3xx23.706=7.902#

Allison Allen · Answer

undefined To find the centroid of a triangle, you can average the x-coordinates and y-coordinates of its vertices. Then, to find the distance from the centroid to the origin, you can use the distance formula.

Centroid coordinates:
$x_{\text{centroid}} = \frac{x_1 + x_2 + x_3}{3}$
$y_{\text{centroid}} = \frac{y_1 + y_2 + y_3}{3}$

Distance from centroid to origin:
$Distance = \sqrt{x_{\text{centroid}}^2 + y_{\text{centroid}}^2}$

Given the coordinates:
$ (7, 9) $, $ (1, 4) $, and $ (3, 8) $

Centroid coordinates:
$x_{\text{centroid}} = \frac{7 + 1 + 3}{3} = \frac{11}{3}$
$y_{\text{centroid}} = \frac{9 + 4 + 8}{3} = \frac{21}{3} = 7$

Distance from centroid to origin:
$Distance = \sqrt{\left(\frac{11}{3}\right)^2 + 7^2}$
$Distance = \sqrt{\frac{121}{9} + 49}$
$Distance = \sqrt{\frac{121 + 441}{9}}$
$Distance = \sqrt{\frac{562}{9}}$
$Distance \approx \sqrt{62.44}$
$Distance \approx 7.9$

So, the distance from the centroid of the triangle to the origin is approximately 7.9 units.