# A triangle has corners at (7 ,6 ), (8 ,2 ), and (5 ,9 ). How far is the triangle's centroid from the origin?

Oct 8, 2016

≈8.75" units"

#### Explanation:

The first step is to establish the position of the centroid.

Given $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$ are the vertices of a triangle, then.

${x}_{c} = \frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right)$ and

${y}_{c} = \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)$

$\Rightarrow {x}_{c} = \frac{1}{3} \left(7 + 8 + 5\right) = \frac{20}{3}$

and ${y}_{c} = \frac{1}{3} \left(6 + 2 + 9\right) = \frac{17}{3}$

$\Rightarrow \text{ coordinates of centroid } = \left(\frac{20}{3} , \frac{17}{3}\right)$

To calculate the distance (d) from the origin to the centroid, use $\textcolor{b l u e}{\text{Pythagoras' theorem}}$

$d = \sqrt{{\left(\frac{20}{3}\right)}^{2} + {\left(\frac{17}{3}\right)}^{2}} = \sqrt{\left(\frac{400}{9}\right) + \left(\frac{289}{9}\right)}$

=sqrt(689/9)≈8.75" to 2 decimal places"