# A triangle has corners at (5 ,3 ), (9 ,7 ), and (6 ,5 ). How far is the triangle's centroid from the origin?

Sep 27, 2016

$\frac{25}{3} \text{ units}$

#### Explanation:

If $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$ are the vertices of a triangle then the coordinates of the $\textcolor{b l u e}{\text{centroid}}$ are.

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

That is the average of the coordinates of the vertices.

Using the given coordinates.

${x}_{c} = \frac{1}{3} \left(5 + 9 + 6\right) = \frac{20}{3} \text{ and } {y}_{c} = \frac{1}{3} \left(3 + 7 + 5\right) = 5$

coordinates of centroid are $\left(\frac{20}{3} , 5\right)$

Since we are calculating the distance ( d) from the origin then, using $\textcolor{b l u e}{\text{Pythagoras' theorem}}$

$d = \sqrt{{\left(\frac{20}{3}\right)}^{2} + {5}^{2}} = \sqrt{\frac{400}{9} + 25}$

$= \sqrt{\frac{400}{9} + \frac{225}{9}} = \sqrt{\frac{625}{9}} = \frac{25}{3}$

distance from centroid to origin =25/3≈8.33" 2 d.p"