# A triangle has corners at (5 ,2 ), (9 ,7 ), and (3 ,5 ). How far is the triangle's centroid from the origin?

Jun 7, 2016

1/3sqrt485≈7.34

#### Explanation:

The first step here is to find the coordinates of the centroid using the following.

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

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

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

using the coordinates given in the question

$\Rightarrow {x}_{c} = \frac{1}{3} \left(5 + 9 + 3\right) = \frac{17}{3}$

and ${y}_{c} = \frac{1}{3} \left(2 + 7 + 5\right) = \frac{14}{3}$

hence coordinates of centroid $= \left(\frac{17}{3} , \frac{14}{3}\right)$

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

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

The 2 points here are (0 ,0) and $\left(\frac{17}{3} , \frac{14}{3}\right)$

$d = \sqrt{{\left(\frac{17}{3} - 0\right)}^{2} + {\left(\frac{14}{3} - 0\right)}^{2}} = \sqrt{{\left(\frac{17}{3}\right)}^{2} + {\left(\frac{14}{3}\right)}^{2}}$

=sqrt(289/9+196/9)=1/3sqrt485≈7.34