A triangle has corners at (-6 ,3 ), (3 ,-2 ), and (5 ,7 ). If the triangle is dilated by a factor of 5  about point #(-2 ,-1 ), how far will its centroid move?

Jun 19, 2017

The distance is $= 18.1$

Explanation:

The centroid of the triangle, $A = \left(- 6 , 3\right)$, $B = \left(3 , - 2\right)$ and $C = \left(5 , 7\right)$ is

${C}_{c} = \left(\frac{- 6 + 3 + 5}{3} , \frac{3 - 2 + 7}{3}\right)$

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

Let the point $D = \left(- 2 , - 1\right)$

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

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

$= 5 < \frac{2}{3} + 2 , \frac{8}{3} + 1 >$

$= < \frac{40}{3} , \frac{55}{3} >$

${C}_{c} ' = \left(\frac{40}{3} , \frac{55}{3}\right) + \left(- 2 , - 1\right)$

$= \left(\frac{34}{3} , \frac{52}{3}\right)$

The distance between the centroids is

$d = \sqrt{{\left(\frac{34}{3} - \frac{2}{3}\right)}^{2} + {\left(\frac{52}{3} - \frac{8}{3}\right)}^{2}}$

$= \sqrt{{\left(\frac{32}{3}\right)}^{2} + {\left(\frac{44}{3}\right)}^{2}}$

$= \frac{1}{3} \sqrt{{32}^{2} + {44}^{2}}$

$= \frac{1}{3} \sqrt{2960}$

$= 18.1$