# A triangle has corners at (5 ,6 ), (2 ,7 ), and (3 ,5 ). How far is the triangle's centroid from the origin?

2/3\sqrt{106}=6.864\ text{unit

#### Explanation:

The coordinates of centroid of given triangle with vertices at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(5 , 6\right)$, $\left({x}_{2} , {y}_{2}\right) \setminus \equiv \left(2 , 7\right)$ & $\left({x}_{3} , {y}_{3}\right) \setminus \equiv \left(3 , 5\right)$ are given as

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

$\setminus \equiv \left(\setminus \frac{5 + 2 + 3}{3} , \setminus \frac{6 + 7 + 5}{3}\right)$

$\setminus \equiv \left(\frac{10}{3} , 6\right)$

hence the distance of centroid of triangle $\left(\frac{10}{3} , 6\right)$ from the origin $\left(0 , 0\right)$ is given by using distance formula

$\setminus \sqrt{{\left(\frac{10}{3} - 0\right)}^{2} + {\left(6 - 0\right)}^{2}}$

$= \frac{2}{3} \setminus \sqrt{106}$

=6.864\ text{unit#