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

Dec 20, 2017

The centroid will move by $= \frac{4}{3} \sqrt{37}$

#### Explanation:

The corners of the triangle are $A = \left(2 , 3\right)$, $B = \left(1 , - 2\right)$ and $C = \left(- 6 , 6\right)$

The centroid of triangle $A B C$ is

$C = \left(\frac{2 + 1 - 6}{3} , \frac{3 - 2 + 6}{3}\right) = \left(- 1 , \frac{7}{3}\right)$

Let the the new centroid be $C ' = \left(x , y\right)$ after dilatation.

Let the fixed point be $D = \left(- 3 , 2\right)$

Let the coefficient of dilatation be $k = 5$

Therefore, in vector notation

$\vec{D C '} = k \vec{D C}$

$\left(\begin{matrix}x - \left(- 3\right) \\ y - 2\end{matrix}\right) = 5 \left(\begin{matrix}- 1 - \left(- 3\right) \\ \frac{7}{3} - 2\end{matrix}\right)$

$x + 3 = 5 \cdot 2$, $\implies$, $x = 7$

$y - 2 = 5 \cdot \left(\frac{1}{3}\right)$, $\implies$, $y = \frac{11}{3}$

The centroid will move by

$= \sqrt{{\left(7 + 1\right)}^{2} + {\left(\frac{11}{3} - \frac{7}{3}\right)}^{2}}$

$= \sqrt{64 + \frac{16}{9}}$

$= \sqrt{\frac{592}{9}}$

$= \frac{4}{3} \sqrt{37}$