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

Jul 28, 2017

The centroid will move by $= 18.1 u$

#### Explanation:

Let the corners of the triangle be $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$

The coordinates of the centroid are

$C = \left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{2} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

Here, we have $\left(8 , 3\right)$, $\left(4 , - 5\right)$, and $\left(2 , 1\right)$

So,

The coordinates of the centroid are $C = \left(\frac{8 + 4 + 2}{3} , \frac{3 - 5 + 1}{3}\right) = \left(\frac{14}{3} , - \frac{1}{3}\right)$

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

The fixed point is $D = \left(1 , - 3\right)$

Therefore,

$\vec{D C '} = 5 \vec{D C}$

$\left(\begin{matrix}x - 1 \\ y + 3\end{matrix}\right) = 5 \left(\begin{matrix}\frac{14}{3} - 1 \\ - \frac{1}{3} + 3\end{matrix}\right) = 5 \left(\begin{matrix}\frac{11}{3} \\ \frac{8}{3}\end{matrix}\right) = \left(\begin{matrix}\frac{55}{3} \\ \frac{40}{3}\end{matrix}\right)$

So,

$x - 1 = \frac{55}{3}$, $\implies$, $x = \frac{55}{3} + 1 = \frac{58}{3}$

$y + 3 = \frac{40}{3}$, $\implies$, $y = \frac{40}{3} - 3 = \frac{31}{3}$

So, the coordinates of $C ' = \left(\frac{58}{3} , \frac{31}{3}\right)$

The distance between the centroids is

$C C ' = \sqrt{{\left(\frac{58}{3} - \frac{14}{3}\right)}^{2} + {\left(\frac{31}{3} + \frac{1}{3}\right)}^{2}}$

$= \sqrt{{\left(\frac{44}{3}\right)}^{2} + {\left(\frac{32}{3}\right)}^{2}}$

$= \frac{\sqrt{{44}^{2} + {32}^{2}}}{3}$

$= 18.1 u$