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

Nov 6, 2017

The triangle's centroid is $8.72$ unit from the origin.

#### Explanation:

Coordinates of the vertices of the triangle are

$A \left(6 , 6\right) , B \left(7 , 4\right) , C \left(5 , 9\right)$. The coordinates of centroid $\left(x , y\right)$ of

triangle is the average of the x-coordinate's value and the average

of the y-coordinate's value of all the vertices of the triangle.

$\therefore x = \frac{6 + 7 + 5}{3} = 6 , y = \frac{6 + 4 + 9}{3} = 6.33 \left(2 \mathrm{dp}\right)$ .

So centroid is at $\left(6 , 6.33\right)$ , Its distance from the origin $\left(0 , 0\right)$

is $D = \sqrt{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}} = \sqrt{{\left(6 - 0\right)}^{2} + {\left(6.33 - 0\right)}^{2}}$ or

$D = \sqrt{{6}^{2} + {6.33}^{2}} = 8.72 \left(2 \mathrm{dp}\right)$ unit.

The triangle's centroid is $8.72 \left(2 \mathrm{dp}\right)$ unit from the origin [Ans]