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

Jun 26, 2016

$\sqrt{52}$

#### Explanation:

Centroid Formula is

$C = \left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$ where

${x}_{1}$, ${x}_{2}$, ${x}_{3}$ are the $x$-coordinates of the vertices of the triangle.
${y}_{1}$, ${y}_{2}$, ${y}_{3}$ are the $y$-coordinate’s of the vertices of the triangle.

In our triangle,

$\left({x}_{1} , {y}_{1}\right) = \left(6 , 6\right)$

$\left({x}_{2} , {y}_{2}\right) = \left(7 , 4\right)$

$\left({x}_{3} , {y}_{3}\right) = \left(5 , 2\right)$

The centroid coordinates are

$C = \left(\frac{6 + 7 + 5}{3} , \frac{6 + 4 + 2}{3}\right) \implies \left(\frac{18}{3} , \frac{12}{3}\right) \implies \left(6 , 4\right)$

Distance from origin $\left(0 , 0\right)$ to $C \left(6 , 4\right)$ using the distance formula is

$D = \sqrt{{6}^{2} + {4}^{2}}$

$= \sqrt{36 + 16}$

$= \sqrt{52}$