# A triangle has corners at (9 ,1 ), (2 ,7 ), and (5 ,8 ). How far is the triangle's centroid from the origin?

Jul 3, 2016

$= \frac{16}{3} \sqrt{2}$

#### Explanation:

Centroid Formula is

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

${x}_{1}$, ${x}_{2}$, ${x}_{3}$ are the $x$-coordinates of the vertices of the triangle.
${y}_{1}$, ${y}_{2}$, ${y}_{3}$ are the $y$-coordinate’s of the vertices of the triangle.

In our triangle,

$\left({x}_{1} , {y}_{1}\right) = \left(9 , 1\right)$

$\left({x}_{2} , {y}_{2}\right) = \left(2 , 7\right)$

$\left({x}_{3} , {y}_{3}\right) = \left(5 , 8\right)$

The centroid coordinates are

$C = \left(\frac{9 + 2 + 5}{3} , \frac{1 + 7 + 8}{3}\right) \implies \left(\frac{16}{3} , \frac{16}{3}\right)$

Distance from origin $\left(0 , 0\right)$ to $C \left(\frac{16}{3} , \frac{16}{3}\right)$ using the distance formula is

$D = \sqrt{{\left(\frac{16}{3}\right)}^{2} + {\left(\frac{16}{3}\right)}^{2}}$

$= \frac{16}{3} \sqrt{2}$