# A triangle has corners at (4 ,3 ), (2 ,6 ), and (7 ,5 ). How far is the triangle's centroid from the origin?

Feb 6, 2017

$\frac{\sqrt{365}}{3} = 6.37$

#### Explanation:

Centroid $\left({x}_{c} , {y}_{c}\right)$ of a triangle with given points $\left(x 1 , y 1\right) , \left(x 2 , y 2\right) \mathmr{and} \left(x 3 , y 3\right)$:
${x}_{c} = \frac{x 1 + x 2 + x 3}{3} ,$
${y}_{c} = \frac{y 1 + y 2 + y 3}{3}$,
where $x 1 , x 2 , x 3 , y 1 , y 2 , y 3$ are the coordinates of the vertices of the triangle.

Given : $A \left(4 , 3\right) , B \left(2 , 6\right) , C \left(7 , 5\right)$
Let $D \left(x , y\right)$ be the centroid of the triangle,

$x = \frac{4 + 2 + 7}{3} = \frac{13}{3}$

$y = \frac{3 + 6 + 5}{3} = \frac{14}{3}$

The distance, $d$, from the origin is :

$d = \sqrt{{\left(\frac{13}{3}\right)}^{2} + {\left(\frac{14}{3}\right)}^{2}} = \sqrt{\frac{365}{9}} = \frac{\sqrt{365}}{3} = 6.37$