# A triangle has corners at (2 ,4 ), (8 ,6 ), and (4 ,9 ). How far is the triangle's centroid from the origin?

Mar 29, 2016

≈ 7.87 units

#### Explanation:

Given the vertices of a triangle A$\left({x}_{1} , {y}_{1}\right) , B \left({x}_{2} , {y}_{2}\right) , C \left({x}_{3} , {y}_{3}\right)$

The centroid has coordinates :

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

For the vertices given :

${x}_{c} = \frac{1}{3} \left(2 + 8 + 4\right) = \frac{14}{3} \text{ and } {y}_{c} = \frac{1}{3} \left(4 + 6 + 9\right) = \frac{19}{3}$

the coordinates of the centroid $= \left(\frac{14}{3} , \frac{19}{3}\right)$

To calculate distance from origin use the $\textcolor{b l u e}{\text{ distance formula }}$

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

since the origin has coordinates (0 , 0) the formula is simplified.

$\Rightarrow d = \sqrt{{\left(\frac{14}{3}\right)}^{2} + {\left(\frac{19}{3}\right)}^{2}}$

 = sqrt(196/9 + 361/9) = sqrt(557/9) ≈ 7.87