# A triangle has corners at (8 ,2 ), (4 ,9 ), and (7 ,3 ). If the triangle is dilated by a factor of 5  about point #(1 ,-6 ), how far will its centroid move?

##### 1 Answer
Mar 17, 2017

THe distance between the centroids is $= 47.7$

#### Explanation:

Let $A B C$ be the triangle

$A = \left(8.2\right)$

$B = \left(4 , 9\right)$

$C = \left(7 , 3\right)$

The centroid of triangle $A B C$ is

${C}_{c} = \left(\frac{8 + 4 + 7}{3} , \frac{2 + 9 + 3}{3}\right) = \left(\frac{19}{3} , \frac{14}{3}\right)$

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

The center of dilatation is $D = \left(1 , - 6\right)$

$\vec{D A '} = 5 \vec{D A} = 5 \cdot < 7 , 8 > = < 35 , 40 >$

$A ' = \left(35 + 1 , 40 - 6\right) = \left(36 , 34\right)$

$\vec{D B '} = 5 \vec{D B} = 5 \cdot < 3 , 15 > = < 15 , 75 >$

$B ' = \left(15 + 1 , 75 - 6\right) = \left(16 , 69\right)$

$\vec{D C '} = 5 \vec{D c} = 5 \cdot < 6 , 9 > = < 30 , 45 >$

$C ' = \left(30 + 1 , 45 - 6\right) = \left(31 , 39\right)$

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

${C}_{c} ' = \left(\frac{36 + 16 + 31}{3} , \frac{34 + 69 + 39}{3}\right) = \left(\frac{83}{3} , \frac{142}{3}\right)$

The distance between the 2 centroids is

${C}_{c} {C}_{c} ' = \sqrt{{\left(\frac{83}{3} - \frac{19}{3}\right)}^{2} + {\left(\frac{142}{3} - \frac{14}{3}\right)}^{2}}$

$= \frac{1}{3} \sqrt{{64}^{2} + {128}^{2}} = \frac{143.1}{3} = 47.7$