# A triangle has corners at (1 ,4 ), (7 ,6 ), and (4 ,5 ). How far is the triangle's centroid from the origin?

##### 1 Answer
Jun 27, 2016

$= \sqrt{41}$

#### 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,

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

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

$\left({x}_{3} , {y}_{3}\right) = \left(4 , 5\right)$

The centroid coordinates are

$C = \left(\frac{1 + 7 + 4}{3} , \frac{4 + 6 + 5}{3}\right) \implies \left(\frac{12}{3} , \frac{15}{3}\right)$ => $\left(4 , 5\right)$

Distance from origin $\left(0 , 0\right)$ to $C \left(4 , 5\right)$ using the distance formula is

$D = \sqrt{{\left(4\right)}^{2} + {\left(5\right)}^{2}}$

$= \sqrt{16 + 25}$

$= \sqrt{41}$