# A triangle has corners at (2 ,4 ), (7 ,6 ), and (4 ,9 ). How far is the triangle's centroid from the origin?

Jun 27, 2016

$= 7.67$

#### 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,

$\left({x}_{1} , {y}_{1}\right) = \left(2 , 4\right)$

$\left({x}_{2} , {y}_{2}\right) = \left(7 , 6\right)$

$\left({x}_{3} , {y}_{3}\right) = \left(4 , 9\right)$

The centroid coordinates are

$C = \left(\frac{2 + 7 + 4}{3} , \frac{4 + 6 + 9}{3}\right) \implies \left(\frac{13}{3} , \frac{19}{3}\right)$

Distance from origin $\left(0 , 0\right)$ to $C \left(\frac{13}{3} , \frac{19}{3}\right)$ using the distance formula is

$D = \sqrt{{\left(\frac{13}{3}\right)}^{2} + {\left(\frac{19}{3}\right)}^{2}}$

$D = \sqrt{{\left(4.33\right)}^{2} + {\left(6.33\right)}^{2}}$

$= \sqrt{18.78 + 40.11}$

$= \sqrt{58.89}$

$= 7.67$