# A triangle has corners at (9 ,3 ), (4 ,4 ), and (3 ,1 ). How far is the triangle's centroid from the origin?

May 25, 2016

Triangle's centroid is $5.963$ 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 $\left(9.3\right)$, $\left(4.4\right)$ and $\left(3 , 1\right)$ is

$\left(\frac{1}{3} \left(9 + 4 + 3\right) , \frac{1}{3} \left(3 + 4 + 1\right)\right)$ or $\left(\frac{16}{3} , \frac{9}{3}\right)$

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

$\sqrt{{\left(\frac{16}{3} - 0\right)}^{2} + {\left(\frac{8}{3} - 0\right)}^{2}} = \sqrt{\frac{256}{9} + \frac{64}{9}}$

= $\frac{1}{3} \sqrt{320} = \frac{8}{3} \times \sqrt{5} = \frac{8}{3} \times 2.236 = 5.963$