# A triangle has corners at (2 ,4 ), (7 ,6 ), and (4 ,5 ). How far is the triangle's centroid from the origin?

Dec 1, 2017

The triangle's centroid is $6.62 \left(2 \mathrm{dp}\right)$ unit from the origin.

#### Explanation:

Coordinates of the vertices of the triangle are

$A \left(2 , 4\right) , B \left(7 , 6\right) , C \left(4 , 5\right)$. 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.

$\therefore x = \frac{2 + 7 + 4}{3} = \frac{13}{3} = 4 \frac{1}{3} , y = \frac{4 + 6 + 5}{3} = 5$ .

So centroid is at $\left(4 \frac{1}{3} , 5\right)$ , Its distance from the origin $\left(0 , 0\right)$

is $D = \sqrt{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}} = \sqrt{{\left(\frac{13}{3} - 0\right)}^{2} + {\left(5 - 0\right)}^{2}}$ or

$D = \sqrt{{\left(\frac{13}{3}\right)}^{2} + {5}^{2}} \approx 6.62 \left(2 \mathrm{dp}\right)$ unit.

The triangle's centroid is $6.62 \left(2 \mathrm{dp}\right)$ unit from the origin [Ans]