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

Dec 21, 2017

The centroid will move by $= 8.6 u$

#### Explanation:

The corners of the triangle are $A = \left(9 , 5\right)$, $B = \left(4 , - 9\right)$ and $C = \left(2 , - 4\right)$

The centroid of triangle $A B C$ is

$C = \left(\frac{9 + 4 + 2}{3} , \frac{5 - 9 - 4}{3}\right) = \left(\frac{15}{3} , - \frac{8}{3}\right)$

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

Let the fixed point be $D = \left(1 , 4\right)$

Let the coefficient of dilatation be $k = \frac{7}{4}$

Therefore, in vector notation

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

$\left(\begin{matrix}x - 1 \\ y - 4\end{matrix}\right) = \frac{7}{4} \left(\begin{matrix}\frac{15}{3} - 1 \\ - \frac{8}{3} - 4\end{matrix}\right)$

$x - 1 = \frac{7}{4} \left(\cdot 4\right)$, $\implies$, $x = 8$

$y - 4 = \frac{7}{4} \cdot \left(- \frac{20}{3}\right)$, $\implies$, $y = - \frac{35}{3} + 4 = - \frac{23}{3}$

The new coordinates of the centroid are $C ' = \left(8 , - \frac{23}{3}\right)$

The centroid will move by

$= \sqrt{{\left(8 - 1\right)}^{2} + {\left(- \frac{23}{3} + \frac{8}{3}\right)}^{2}}$

$= \sqrt{49 + 25}$

$= \sqrt{74}$

$= 8.6 u$