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

Apr 30, 2016

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

#### Explanation:

Coordinates of centroid of a triangle whose vertices (corners) are $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$ is given by

$\left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

As corners of triangle are $\left(5 , 2\right)$, $\left(2 , 7\right)$ and $\left(3 , 5\right)$, the centroid of given triangle is $\left(\frac{5 + 2 + 3}{3} , \frac{2 + 7 + 5}{3}\right)$ or $\left(\frac{10}{3} , \frac{14}{3}\right)$.

And its distance from origin is $\sqrt{{\left(\frac{10}{3} - 0\right)}^{2} + {\left(\frac{14}{3} - 0\right)}^{2}} = \sqrt{\frac{100}{9} + \frac{196}{9}} = \sqrt{\frac{296}{9}}$

= $\frac{1}{3} \sqrt{2} \times 2 \times 74 = \frac{2}{3} \times \sqrt{74} = \frac{2}{3} \times 8.602 = 5.735$