# A triangle has corners at (5 ,6 ), (3 ,7 ), and (8 ,9 ). How far is the triangle's centroid from the origin?

Feb 3, 2017

The answer is $= 9.1$

#### Explanation:

The centroid$\left({x}_{c} , {y}_{c}\right)$ of a triangle with corners $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$ is

${x}_{c} = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{3}$

${y}_{c} = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}$

Therefore,

${x}_{c} = \frac{5 + 3 + 8}{3} = \frac{16}{3}$

${y}_{c} = \frac{6 + 7 + 9}{3} = \frac{22}{3}$

The distance from the origin is

$= \sqrt{{x}_{c}^{2} + {y}_{c}^{2}}$

$= \sqrt{{16}^{2} / 9 + {22}^{2} / 9}$

$= \frac{\sqrt{{16}^{2} + {22}^{2}}}{3}$

$= \frac{\sqrt{740}}{3} = 9.1$