# A triangle has corners at (-1 ,7 ), (-5 ,-3 ), and (2 ,9 ). If the triangle is dilated by a factor of 5  about point #(-7 ,1 ), how far will its centroid move?

Mar 20, 2017

The distance is $= 24$

#### Explanation:

Let $A B C$ be the triangle

$A = \left(- 1 , 7\right)$

$B = \left(- 5 , - 3\right)$

$C = \left(2 , 9\right)$

The centroid of triangle $A B C$ is

${C}_{c} = \left(\frac{- 1 - 5 + 2}{3} , \frac{7 + \left(- 3\right) + 9}{3}\right) = \left(\frac{4}{3} , \frac{13}{3}\right)$

Let $A ' B ' C '$ be the triangle after the dilatation

The center of dilatation is $D = \left(- 7 , 1\right)$

$\vec{D A '} = 5 \vec{D A} = 5 \cdot < 6 , 6 > = < 30 , 30 >$

$A ' = \left(30 - 7 , 30 + 1\right) = \left(23 , 31\right)$

$\vec{D B '} = 5 \vec{D B} = 5 \cdot < 2 , - 4 > = < 10 , - 20 >$

$B ' = \left(10 - 7 , - 20 + 1\right) = \left(3 , - 19\right)$

$\vec{D C '} = 5 \vec{D C} = 5 \cdot < 9 , 8 > = < 45 , 40 >$

$C ' = \left(45 - 7 , 40 + 1\right) = \left(38 , 41\right)$

The centroid ${C}_{c} '$ of triangle $A ' B ' C '$ is

${C}_{c} ' = \left(\frac{23 + 3 + 38}{3} , \frac{31 - 19 + 41}{3}\right) = \left(\frac{64}{3} , \frac{53}{3}\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(\frac{64}{3} - \frac{4}{3}\right)}^{2} + {\left(\frac{53}{3} - \frac{13}{3}\right)}^{2}}$

$= \frac{1}{3} \sqrt{{60}^{2} + {40}^{2}} = \frac{72.11}{3} = 24$