# A triangle has corners at (2 ,3 ), (4 ,-1 ), and (-8 ,6 ). If the triangle is dilated by a factor of 5  about point #(4 ,-3 ), how far will its centroid move?

##### 1 Answer
Jul 3, 2017

The centroid will move by $= 33.4$

#### Explanation:

The centroid of the triangle, $A = \left(2 , 3\right)$, $B = \left(4 , - 1\right)$ and $C = \left(- 8 , 6\right)$ is

${C}_{c} = \left(\frac{2 + 4 - 8}{3} , \frac{3 - 1 + 6}{3}\right)$

$= \left(- \frac{2}{3} , \frac{8}{3}\right)$

Let the point $D = \left(4 , - 3\right)$

Let the centroid be ${C}_{c} '$ after the dilatation

$\vec{D {C}_{c} '} = 4 \vec{D {C}_{c}}$

$= 5 < - \frac{2}{3} - 4 , \frac{8}{3} + 3 >$

$= < - \frac{70}{3} , \frac{85}{3} >$

${C}_{c} ' = \left(- \frac{70}{3} , \frac{85}{3}\right) + \left(4 , - 3\right)$

$= \left(- \frac{58}{3} , \frac{74}{3}\right)$

The distance between the centroids is

$d = \sqrt{{\left(- \frac{58}{3} + \frac{2}{3}\right)}^{2} + {\left(\frac{74}{3} + 3\right)}^{2}}$

$= \sqrt{{\left(\frac{56}{3}\right)}^{2} + {\left(\frac{83}{3}\right)}^{2}}$

$= \frac{\sqrt{10025}}{3}$

$= 33.4$