# A triangle has corners at (6 ,3 ), (2 ,6 ), and (1 ,5 ). How far is the triangle's centroid from the origin?

Jul 13, 2018

$\text{Distance} = \frac{\sqrt{277}}{3}$

#### Explanation:

The centroid of a triangle with vertices at $\left({x}_{A} , {y}_{A}\right)$, $\left({x}_{B} , {y}_{B}\right)$ and $\left({x}_{C} , {y}_{C}\right)$ has coordinates

$O \left(\frac{{x}_{A} + {x}_{B} + {x}_{C}}{3} , \frac{{y}_{A} + {y}_{B} + {y}_{C}}{3}\right)$

Hence:

$\left\{\begin{matrix}A \left(6 3\right) \\ B \left(2 6\right) \\ C \left(1 5\right)\end{matrix}\right. \implies O \left(\frac{6 + 2 + 1}{3} , \frac{3 + 6 + 5}{3}\right) \equiv O \left(3 , \frac{14}{3}\right)$

The distance between two points $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ is given by the formula

$\text{Distance} = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

The distance between the centroid, $O$, and the origin, $\left(0 , 0\right)$ is going to be:

$\text{Distance} = \sqrt{{\left(\textcolor{red}{3} - 0\right)}^{2} + {\left(\textcolor{red}{\frac{14}{3}} - 0\right)}^{2}}$

$= \sqrt{9 + \frac{196}{9}} = \sqrt{\frac{277}{9}} = \frac{\sqrt{227}}{3}$