A triangle has corners at $(4 ,-6 )$ , $(3 ,2 )$ , and $(1 ,3 )$ . If the triangle is dilated by a factor of $1/3$ about point #(6 ,2 ), how far will its centroid move?

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Answer 1 · 2017-04-26T10:31:27

The distance is $=2.48$

Let $ABC$ be the triangle

$A=(4,-6)$

$B=(3,2)$

$C=(1,3)$

The centroid of triangle $ABC$ is

$C_c=((4+3+1)/3,((-6)+2+3)/3)=(8/3,-1/3)$

Let $A'B'C'$ be the triangle after the dilatation

The center of dilatation is $D=(6,-2)$

$vec(DA')=1/3vec(DA)=1/3*<-2,-4> = <-2/3,-4/3>$

$A'=(-2/3+6,-4/3-2)=(16/3,-10/3)$

$vec(DB')=1/3vec(DB)=1/3*<-3,4> = <-1,4/3>$

$B'=(-1+6,4/3-2)=(5,-2/3)$

$vec(DC')=1/3vec(DC)=1/3*<-5,5> = <-5/3,5/3>$

$C'=(-5/3+6,5/3-2)=(13/3,-1/3)$

The centroid $C_c'$ of triangle $A'B'C'$ is

$C_c'=((16/3+5+13/3)/3,(-10/3-2/3-1/3)/3)=(44/9,-13/9)$

The distance between the 2 centroids is

$C_cC_c'=sqrt((44/9-8/3)^2+(-13/9+1/3)^2)$

$=1/9sqrt(20^2+10^2)=2.48$

We can do this directly

$vecDC'_c=1/3vecDC_c=1/3(8/3-6,-1/3+2)=(-10/9,5/9)$

$C'_c=(-10/9+6,5/9-2)=(44/9,-13/9)$

A triangle has corners at (4 ,-6 )(4 ,-6 ), (3 ,2 )(3 ,2 ), and (1 ,3 )(1 ,3 ). If the triangle is dilated by a factor of 1/3 1/3 about point #(6 ,2 ), how far will its centroid move?