# A triangle has corners at (9 ,5 ), (2 ,7 ), and (3 ,4 ). How far is the triangle's centroid from the origin?

Sep 26, 2017

$\frac{2}{3} \sqrt{103}$

#### Explanation:

a triangle with vertices at $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$ has centroid at $\left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

A triangle has corners at (9,5), (2,7) and (3,4)

Hence centroid is $\left(\frac{9 + 2 + 3}{3} , \frac{5 + 7 + 4}{3}\right)$
or $\left(\frac{14}{3} , \frac{16}{3}\right)$

distance between centroid$\left(\frac{14}{3} , \frac{16}{3}\right)$ and origin $\left(0 , 0\right)$
distance formula d=sqrt((x_2-x_1)^2+(y_2-y_1)^2
d=sqrt((14/3-0)^2+(16/3-0)^2
$d = \sqrt{\frac{196}{9} + \frac{216}{9}} = \sqrt{\frac{412}{9}} = \frac{2}{3} \sqrt{103}$