# A triangle has corners at (4 ,1 ), (6 ,3 ), and (3 ,8 ). How far is the triangle's centroid from the origin?

Apr 29, 2016

Distance of centroid from origin is $5.897$

#### Explanation:

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

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

As corners of triangle are $\left(4 , 1\right)$, $\left(6 , 3\right)$ and $\left(3 , 8\right)$, the centroid of given triangle is $\left(\frac{4 + 6 + 3}{3} , \frac{1 + 3 + 8}{3}\right)$ or $\left(\frac{13}{3} , 4\right)$.

And its distance from origin is $\sqrt{{\left(\frac{13}{3} - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{\frac{169}{9} + 16} = \sqrt{\frac{169}{9} + \frac{144}{9}}$

= $\sqrt{\frac{169 + 144}{9}} = \sqrt{\frac{313}{9}} = \frac{1}{3} \sqrt{313} = \frac{1}{3} \times 17.692 = 5.897$