# A triangle has corners at (1 ,6 ), (8 ,2 ), and (5 ,9 ). How far is the triangle's centroid from the origin?

Nov 9, 2016

Triangle's centroid is $7.341$ units away from the origin.

#### Explanation:

Centroid of a triangle, whose corners are $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$, is given by $\left(\frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right) , \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)\right)$

Hence centroid of the triangle whose corners are $\left(1 , 6\right)$, $\left(8 , 2\right)$ and $\left(5 , 9\right)$ is

$\left(\frac{1}{3} \left(1 + 8 + 5\right) , \frac{1}{3} \left(6 + 2 + 9\right)\right)$ or $\left(\frac{14}{3} , \frac{17}{3}\right)$

And its distance from origin $\left(0 , 0\right)$ is

$\sqrt{{\left(\frac{14}{3} - 0\right)}^{2} + {\left(\frac{17}{3} - 0\right)}^{2}} = \sqrt{\frac{196}{9} + \frac{289}{9}}$

= $\frac{1}{3} \sqrt{485} = \frac{1}{3} \times 22.023 = 7.341$