# A triangle has corners at (7 ,9 ), (1 ,1 ), and (3 ,8 ). How far is the triangle's centroid from the origin?

Sep 6, 2016

Triangle's centroid is $7.032$ 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 $\left(7 , 9\right)$, $\left(1.1\right)$ and $\left(3 , 8\right)$ is

$\left(\frac{1}{3} \left(7 + 1 + 3\right) , \frac{1}{3} \left(9 + 1 + 8\right)\right)$ or $\left(\frac{11}{3} , \frac{18}{3}\right)$

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

$\sqrt{{\left(\frac{18}{3} - 0\right)}^{2} + {\left(\frac{11}{3} - 0\right)}^{2}} = \sqrt{\frac{324}{9} + \frac{121}{9}}$

= $\sqrt{\frac{445}{9}} = \frac{1}{3} \sqrt{445} = \frac{1}{3} \times 21.095 = 7.032$