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

May 9, 2017

The centroid will move by $= 33.9$

#### Explanation:

Let $A B C$ be the triangle

$A = \left(2 , 3\right)$

$B = \left(1 , - 2\right)$

$C = \left(- 6 , 6\right)$

The centroid of triangle $A B C$ is

${C}_{c} = \left(\frac{2 + 1 - 6}{3} , \frac{3 - 2 + 6}{3}\right) = \left(- \frac{3}{3} , \frac{7}{3}\right) = \left(- 1 , \frac{7}{3}\right)$

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

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

Then,

$\vec{D {C}_{c} '} = 5 \vec{D {C}_{c}} = 5 \left(- 1 - 4 , \frac{7}{3} + 3\right) = \left(- 25 , \frac{80}{3}\right)$

${C}_{c} ' = \left(- 25 , \frac{80}{3}\right) + \left(4 , 3\right) = \left(- 21 , \frac{89}{3}\right)$

Therefore,

the distance ${C}_{c} {C}_{c} '$

$= \sqrt{{\left(- 21 + 1\right)}^{2} + {\left(\frac{89}{3} - \frac{7}{3}\right)}^{2}}$

$= \sqrt{400 + 747.1}$

$= 33.9$