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

May 12, 2017

The centroid will move by $= \frac{\sqrt{68}}{3} = 2.75$

#### Explanation:

The centroid of the triangle, $A = \left(5 , 2\right)$, $B = \left(- 3 , - 3\right)$ and $C = \left(1 , - 2\right)$ is

${C}_{c} = \left(\frac{5 - 3 + 1}{3} , \frac{2 - 3 - 2}{3}\right)$

$= \left(1 , - 1\right)$

Let the point $D = \left(5 , - 2\right)$

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

$\vec{D {C}_{c} '} = \frac{1}{3} \vec{D {C}_{c}}$

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

$= < - \frac{4}{3} , \frac{1}{3} >$

${C}_{c} ' = \left(- \frac{4}{3} , \frac{1}{3}\right) + \left(5 , - 2\right)$

$= \left(\frac{11}{3} , - \frac{5}{3}\right)$

The distance between the centroids is

$d = \sqrt{{\left(\frac{11}{3} - 1\right)}^{2} + {\left(- \frac{5}{3} + 1\right)}^{2}}$

$= \sqrt{\frac{64}{9} + \frac{4}{9}}$

$= \frac{\sqrt{68}}{3}$

$= 2.75$