# A triangle has corners at (9 ,3 ), (7 ,4 ), and (3 ,1 ). How far is the triangle's centroid from the origin?

May 25, 2016

Triangle's centroid from the origin is $6.872$ units away.

#### 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 $\left(9 , 3\right)$, $\left(7.4\right)$ and $\left(3 , 1\right)$ is

$\left(\frac{1}{3} \left(9 + 7 + 3\right) , \frac{1}{3} \left(3 + 4 + 1\right)\right)$ or $\left(\frac{19}{3} , \frac{8}{3}\right)$

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

$\sqrt{{\left(\frac{19}{3} - 0\right)}^{2} + {\left(\frac{8}{3} - 0\right)}^{2}} = \sqrt{\frac{361}{9} + \frac{64}{9}}$

= $\frac{1}{3} \sqrt{425} = \frac{1}{3} \times 20.616 = 6.872$

May 25, 2016

Triangle's centroid is $6.872$ 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 $\left(9 , 3\right)$, $\left(7.4\right)$ and $\left(3 , 1\right)$ is

$\left(\frac{1}{3} \left(9 + 7 + 3\right) , \frac{1}{3} \left(3 + 4 + 1\right)\right)$ or $\left(\frac{19}{3} , \frac{8}{3}\right)$

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

$\sqrt{{\left(\frac{19}{3} - 0\right)}^{2} + {\left(\frac{8}{3} - 0\right)}^{2}} = \sqrt{\frac{361}{9} + \frac{64}{9}}$

= $\frac{1}{3} \sqrt{425} = \frac{1}{3} \times 20.616 = 6.872$