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

Jun 10, 2017

The distance is $= 4.62$

Explanation:

Let the centroid of the triangle be $C$

Then,

$C = \left(\frac{9 - 2 + 3}{3} , \frac{- 5 - 1 - 4}{3}\right)$

$= \left(\frac{10}{3} , - \frac{10}{3}\right)$

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

The point $D = \left(1 , 4\right)$

So,

$\vec{D C '} = \frac{2}{5} \vec{D C}$

$\left(\begin{matrix}x ' - 1 \\ y ' - 4\end{matrix}\right) = \frac{2}{5} \left(\begin{matrix}\frac{10}{3} - 1 \\ - \frac{10}{3} - 4\end{matrix}\right)$

$x ' - 1 = \frac{2}{5} \left(\frac{10}{3} - 1\right) = \frac{14}{15}$

$\implies$, $x ' = \frac{14}{15} + 1 = \frac{29}{15}$

$y ' - 4 = \frac{2}{5} \left(- \frac{10}{3} - 4\right) = - \frac{44}{15}$

$y ' = - \frac{44}{15} + 4 = \frac{16}{15}$

Therefore,

The new centroid is $C ' = \left(\frac{29}{15} , \frac{16}{15}\right)$

The distance

$\mathbb{C} ' = \sqrt{{\left(\frac{10}{3} - \frac{29}{15}\right)}^{2} + {\left(- \frac{10}{3} - \frac{16}{15}\right)}^{2}}$

$= \sqrt{{\left(\frac{21}{15}\right)}^{2} + {\left(\frac{66}{15}\right)}^{2}}$

$= \frac{1}{15} \sqrt{{21}^{2} + {66}^{2}}$

$= 4.62$