# A triangle has corners at (4 ,6 ), (8 ,3 ), and (3 ,5 ). How far is the triangle's centroid from the origin?

May 4, 2016

Approximately $6.84$ units

#### Explanation:

In general , given a triangle's corners at:
$\textcolor{w h i t e}{\text{XXX}} P = \left({P}_{x} , {P}_{y}\right)$
$\textcolor{w h i t e}{\text{XXX}} Q = \left({Q}_{x} , {Q}_{y}\right)$
$\textcolor{w h i t e}{\text{XXX}} R = \left({R}_{x} , {R}_{y}\right)$
the centroid is located at
$\textcolor{w h i t e}{\text{XXX}} C = \left(\frac{{P}_{x} + {Q}_{x} + {r}_{x}}{3} , \frac{{P}_{y} + {Q}_{y} + {R}_{y}}{3}\right)$

In this case with corners $\left(4 , 6\right) , \left(8 , 3\right) , \mathmr{and} \left(3 , 5\right)$
the centroid is at
$\textcolor{w h i t e}{\text{XXX}} \left(\frac{4 + 8 + 3}{3} , \frac{6 + 3 + 5}{3}\right) = \left(5 , \frac{14}{3}\right)$

The distance from the origin (i.e. from $\left(0 , 0\right)$) is given by the Pythagorean Theorem
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{\left(5 - 0\right)}^{2} + {\left(\frac{14}{3}\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{25 \times 9 + 196}}{3}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{421}}{3}$

$\textcolor{w h i t e}{\text{XXX}} \approx 6.84$