# A triangle has corners at (2 ,2 ), (4 ,-7 ), and (3 ,4 ). If the triangle is dilated by a factor of 5  about point #(1 ,-9 ), how far will its centroid move?

Mar 30, 2017

The centroid will move by $= 8.69$

#### Explanation:

Let $A B C$ be the triangle

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

$B = \left(4 , - 7\right)$

$C = \left(3 , 4\right)$

The centroid of triangle $A B C$ is

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

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

The center of dilatation is $D = \left(1 , - 9\right)$

$\vec{D A '} = 5 \vec{D A} = 5 \cdot < 1 , 11 > = < 5 , 55 >$

$A ' = \left(5 + 1 , 55 - 9\right) = \left(6 , 46\right)$

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

$B ' = \left(15 + 1 , 10 - 9\right) = \left(16 , 1\right)$

$\vec{D C '} = 5 \vec{D C} = 5 \cdot < 2 , 13 > = < 10 , 65 >$

$C ' = \left(10 + 1 , 65 - 9\right) = \left(11 , 56\right)$

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

${C}_{c} ' = \left(\frac{6 + 16 + 11}{3} , \frac{46 + 1 + 56}{3}\right) = \left(11 , \frac{103}{3}\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(11 - 3\right)}^{2} + {\left(\frac{103}{3} + \frac{1}{3}\right)}^{2}}$

$= \frac{1}{3} \sqrt{64 \cdot 9 + {104}^{2}} = \frac{26.08}{3} = 8.69$