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

Jun 12, 2017

The centroid will move by $= 21.3$

#### Explanation:

The centroid of the triangle, $A = \left(3 , 5\right)$, $B = \left(5 , 2\right)$ and $C = \left(9 , 4\right)$ is

${C}_{c} = \left(\frac{3 + 5 + 9}{3} , \frac{5 + 2 + 4}{3}\right)$

$= \left(\frac{17}{3} , \frac{11}{3}\right)$

Let the point $D = \left(1 , 9\right)$

Let the centroid be ${C}_{c} '$ after the dilatation

$\vec{D {C}_{c} '} = 4 \vec{D {C}_{c}}$

$= 4 < \frac{17}{3} - 1 , \frac{11}{3} - 9 >$

$= < \frac{56}{3} , - \frac{64}{3} >$

${C}_{c} ' = \left(\frac{56}{3} , - \frac{64}{3}\right) + \left(1 , 9\right)$

$= \left(\frac{59}{3} , - \frac{37}{3}\right)$

The distance between the centroids is

$d = \sqrt{{\left(\frac{59}{3} - \frac{17}{3}\right)}^{2} + {\left(- \frac{37}{3} - \frac{11}{3}\right)}^{2}}$

$= \sqrt{{\left(\frac{42}{3}\right)}^{2} + {\left(\frac{48}{3}\right)}^{2}}$

$= \sqrt{{14}^{2} + {16}^{2}}$

$= \sqrt{452}$

$= 21.3$