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

Feb 26, 2018

Distance moved by centroid color(green)(vec(GG’) = 0.9068

#### Explanation:

Given : $A \left(2 , - 5\right) , B \left(- 8 , 4\right) , C \left(1 , - 3\right)$ Dilated about $D \left(- 1 , - 2\right)$ by a factor of $\frac{2}{5}$

To find how far the centroid has moved.

Centroid $G \left(x , y\right) = \frac{2 - 8 + 1}{3} , \frac{- 5 + 4 - 3}{3} \implies \left(- \frac{5}{3} , - \frac{4}{3}\right)$

vec(A’D) = (2/5) vec(AD)

a’((x),(y)) - d((-1),(-2)) = (2/5)( a((2),(-5)) - d((-1),(-2)))

a’((x),(y)) = (2/5) ((2),(-5)) - (2/5)((-1),(-2)) + ((-1),(-2))

$\implies \left(\frac{2}{5}\right) \left(\begin{matrix}2 \\ - 5\end{matrix}\right) + \left(\frac{3}{5}\right) \left(\begin{matrix}- 1 \\ - 2\end{matrix}\right) = \left(\begin{matrix}\frac{4}{5} \\ - 2\end{matrix}\right) + \left(\begin{matrix}- \frac{3}{5} \\ - \frac{6}{5}\end{matrix}\right)$

a’((x),(y)) = ((1/5),(-16/5))

Similarly, b’((x),(y)) = (2/5)((-8),(4)) + (3/5)((-1),(-2)) = ((-16/5),(8/5)) + ((-3/5),(-6/5))

b’((x),(y)) = ((-19/5),(2/5))

Similarly, c’((x),(y)) = (2/5)((1),(-3)) + (3/5)((-1),(-2)) = ((2/5),(-3/5)) + ((-3/5),(-6/5))

c’((x),(y)) = ((-1/5),(-9/5))

G’(x,y) = (-3/5 -19/5 - 1/5)/3, (-16/5 + 2/5 - 9/5)/3) => (-23/15, —23/15)

Distance moved by centroid

vec(GG’) = sqrt((-5/3 + 23/15)^2 + (-4/3+23/15)^2) = 0.9068#