# A triangle has corners at (1 ,6 ), (7 ,4 ), and (5 ,9 ). How far is the triangle's centroid from the origin?

the centroid is $58.89$ when 3 vertices of a triangle (xunits away from the origin.
when 3 vertices of a triangle are $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$ then the centroid of the triangle is $G = \left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$
substituting the values we get the centroid as $G = \left(\frac{13}{3} , \frac{19}{3}\right)$
the distance between $G \left(\frac{13}{3} , \frac{19}{3}\right) \mathmr{and} C \left(0 , 0\right)$ is sqrt[(x_2-x_1)^2+(y_2-y_1)^2# substituting in the formula we get distance as $\sqrt{\frac{530}{9}} \approx 58.89$