# A triangle has corners at (5 ,6 ), (3 ,2 ), and (8 ,9 ). How far is the triangle's centroid from the origin?

Nov 26, 2016

The distance is $= \frac{\sqrt{545}}{3} = 7.8$

#### Explanation:

Let the vertices of a triangle be $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and
$\left({x}_{3} , {y}_{3}\right)$

Then the centroid is $\left({x}_{c} , {y}_{c}\right)$

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

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

And the distance of the centroid from the origin is

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

Here we have, $\left(5 , 6\right)$, $\left(3 , 2\right)$ and $\left(8 , 9\right)$

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

${y}_{c} = \frac{6 + 2 + 9}{3} = \frac{17}{3}$

$d = \sqrt{{16}^{2} / {3}^{2} + {17}^{2} / {3}^{2}}$

$= \frac{\sqrt{545}}{3} = 7.8$