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

Apr 26, 2017

The distance is $= 2.48$

#### Explanation:

Let $A B C$ be the triangle

$A = \left(4 , - 6\right)$

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

$C = \left(1 , 3\right)$

The centroid of triangle $A B C$ is

${C}_{c} = \left(\frac{4 + 3 + 1}{3} , \frac{\left(- 6\right) + 2 + 3}{3}\right) = \left(\frac{8}{3} , - \frac{1}{3}\right)$

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

The center of dilatation is $D = \left(6 , - 2\right)$

$\vec{D A '} = \frac{1}{3} \vec{D A} = \frac{1}{3} \cdot < - 2 , - 4 > = < - \frac{2}{3} , - \frac{4}{3} >$

$A ' = \left(- \frac{2}{3} + 6 , - \frac{4}{3} - 2\right) = \left(\frac{16}{3} , - \frac{10}{3}\right)$

$\vec{D B '} = \frac{1}{3} \vec{D B} = \frac{1}{3} \cdot < - 3 , 4 > = < - 1 , \frac{4}{3} >$

$B ' = \left(- 1 + 6 , \frac{4}{3} - 2\right) = \left(5 , - \frac{2}{3}\right)$

$\vec{D C '} = \frac{1}{3} \vec{D C} = \frac{1}{3} \cdot < - 5 , 5 > = < - \frac{5}{3} , \frac{5}{3} >$

$C ' = \left(- \frac{5}{3} + 6 , \frac{5}{3} - 2\right) = \left(\frac{13}{3} , - \frac{1}{3}\right)$

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

${C}_{c} ' = \left(\frac{\frac{16}{3} + 5 + \frac{13}{3}}{3} , \frac{- \frac{10}{3} - \frac{2}{3} - \frac{1}{3}}{3}\right) = \left(\frac{44}{9} , - \frac{13}{9}\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(\frac{44}{9} - \frac{8}{3}\right)}^{2} + {\left(- \frac{13}{9} + \frac{1}{3}\right)}^{2}}$

$= \frac{1}{9} \sqrt{{20}^{2} + {10}^{2}} = 2.48$

We can do this directly

$\vec{D} C {'}_{c} = \frac{1}{3} \vec{D} {C}_{c} = \frac{1}{3} \left(\frac{8}{3} - 6 , - \frac{1}{3} + 2\right) = \left(- \frac{10}{9} , \frac{5}{9}\right)$

$C {'}_{c} = \left(- \frac{10}{9} + 6 , \frac{5}{9} - 2\right) = \left(\frac{44}{9} , - \frac{13}{9}\right)$