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

Mar 21, 2017

The distance is $= 29$

#### Explanation:

Let $A B C$ be the triangle

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

$B = \left(6 , - 3\right)$

$C = \left(- 1 , 4\right)$

The centroid of triangle $A B C$ is

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

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

The center of dilatation is $D = \left(4 , - 6\right)$

$\vec{D A '} = 5 \vec{D A} = 5 \cdot < - 6 , 7 > = < - 30 , 35 >$

$A ' = \left(- 30 + 4 , 35 - 6\right) = \left(- 26 , 29\right)$

$\vec{D B '} = 5 \vec{D B} = 5 \cdot < 2 , 3 > = < 10 , 15 >$

$B ' = \left(10 + 4 , 15 - 6\right) = \left(14 , 9\right)$

$\vec{D C '} = 5 \vec{D C} = 5 \cdot < - 5 , 10 > = < - 25 , 50 >$

$C ' = \left(- 25 + 4 , 50 - 6\right) = \left(- 19 , 44\right)$

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

${C}_{c} ' = \left(\frac{- 26 + 14 - 19}{3} , \frac{29 + 9 + 44}{3}\right) = \left(- \frac{31}{3} , \frac{82}{3}\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(- \frac{31}{3} - 1\right)}^{2} + {\left(\frac{82}{3} - \frac{2}{3}\right)}^{2}}$

$= \frac{1}{3} \sqrt{{34}^{2} + {80}^{2}} = \frac{86.93}{3} = 29$