# A triangle has corners at (1 ,6 ), (3 ,2 ), and (8 ,9 ). How far is the triangle's centroid from the origin?

\sqrt{433}/3=6.936\ \text{unit

#### Explanation:

Given that the vertices of a triangle are $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(1 , 6\right)$, $\left({x}_{2} , {y}_{2}\right) \setminus \equiv \left(3 , 2\right)$ & $\left({x}_{3} , {y}_{3}\right) \setminus \equiv \left(8 , 9\right)$ then the coordinates of centroid of triangle are given as

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

$\setminus \equiv \left(\setminus \frac{1 + 3 + 8}{3} , \setminus \frac{6 + 2 + 9}{3}\right)$

$\setminus \equiv \left(4 , \setminus \frac{17}{3}\right)$

hence the distance between the centroid $\left(4 , \frac{17}{3}\right)$ & the origin $\left(0 , 0\right)$ is given by distance formula as follows

$\setminus \sqrt{{\left(4 - 0\right)}^{2} + {\left(\frac{17}{3} - 0\right)}^{2}}$

$= \setminus \frac{\sqrt{433}}{3}$

=6.936\ \text{unit