# A triangle has corners at (1 ,9 ), (5 ,4 ), and (6 ,2 ). How far is the triangle's centroid from the origin?

$\sqrt{41}$ units
A triangle with vertices at $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \mathmr{and} \left({x}_{3} , {y}_{3}\right)$ has centroid at $\left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$
$\implies$ centroid of the triangle $= \left(\frac{1 + 5 + 6}{3} , \frac{9 + 4 + 2}{3}\right) = \left(4 , 5\right)$
Let $d$ be the distance between the centroid and the origin $O \left(0 , 0\right)$,
$\implies d = \sqrt{{4}^{2} + {5}^{2}} = \sqrt{16 + 25} = \sqrt{41}$ units