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

Mar 14, 2017

The distance is $= 5.83$

#### Explanation:

Let $A B C$ be the triangle

$A = \left(5.3\right)$

$B = \left(1 , 4\right)$

$C = \left(3.5\right)$

The centroid of triangle $A B C$ is

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

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

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

$\vec{D A '} = 3 \vec{D A} = 3 \cdot < 3 , 1 > = < 9 , 3 >$

$A ' = \left(9 + 2 , 3 + 2\right) = \left(11 , 5\right)$

$\vec{D B '} = 3 \vec{D B} = 3 \cdot < - 1 , 2 > = < - 3 , 6 >$

$B ' = \left(- 3 + 2 , 6 + 2\right) = \left(- 1 , 11\right)$

$\vec{D C '} = 3 \vec{D c} = 3 \cdot < 2 , 3 > = < 6 , 9 >$

$C ' = \left(6 + 2 , 9 + 2\right) = \left(8 , 11\right)$

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

${C}_{c} ' = \left(\frac{11 - 1 + 8}{3} , \frac{5 + 11 + 11}{3}\right) = \left(6 , 9\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(6 - 3\right)}^{2} + {\left(9 - 4\right)}^{2}}$

$= \sqrt{9 + 25} = \sqrt{34} = 5.83$