# A triangle has corners at (4 ,6 ), (3 ,9 ), and (6 ,5 ). How far is the triangle's centroid from the origin?

Sep 1, 2017

The triangle's centroid is $7.95$ unit far from the origin $\left(0 , 0\right)$

#### Explanation:

Corners of triangle co-ordinates are

$A \left({x}_{1} = 4 , {y}_{1} = 6\right) , B \left({x}_{2} = 3 , {y}_{2} = 9\right) , C \left({x}_{3} = 6 , {y}_{3} = 5\right)$

We know centroid $\left(x\right) = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} = \frac{4 + 3 + 6}{3} = \frac{13}{3}$ and

centroid $\left(y\right) = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3} = \frac{6 + 9 + 5}{3} = \frac{20}{3}$

So centroid co-ordinates are $\left(x = \frac{13}{3} , y = \frac{20}{3}\right)$

Distance of centroid from origin$\left(0 , 0\right)$ is

$D = \sqrt{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}} = \sqrt{{\left(\frac{13}{3}\right)}^{2} + {\left(\frac{20}{3}\right)}^{2}} \approx 7.95$ unit

The triangle's centroid is $7.95$ unit far from the origin $\left(0 , 0\right)$ [Ans]