A triangle has corners at $(-2 ,1 )$ , $(6 ,-3 )$ , and $(-1 ,4 )$ . If the triangle is dilated by a factor of $5$ about point #(4 ,-6 ), how far will its centroid move?

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Answer 1 · 2017-03-21T13:11:47

The distance is $=29$

Let $ABC$ be the triangle

$A=(-2,1)$

$B=(6,-3)$

$C=(-1,4)$

The centroid of triangle $ABC$ is

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

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

The center of dilatation is $D=(4,-6)$

$vec(DA')=5vec(DA)=5*<-6,7> = <-30,35>$

$A'=(-30+4,35-6)=(-26,29)$

$vec(DB')=5vec(DB)=5*<2,3> = <10,15>$

$B'=(10+4,15-6)=(14,9)$

$vec(DC')=5vec(DC)=5*<-5,10> = <-25,50>$

$C'=(-25+4,50-6)=(-19,44)$

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

$C_c'=((-26+14-19)/3,(29+9+44)/3)=(-31/3,82/3)$

The distance between the 2 centroids is

$C_cC_c'=sqrt((-31/3-1)^2+(82/3-2/3)^2)$

$=1/3sqrt(34^2+80^2)=86.93/3=29$

A triangle has corners at (-2 ,1 )(-2 ,1 ), (6 ,-3 )(6 ,-3 ), and (-1 ,4 )(-1 ,4 ). If the triangle is dilated by a factor of 5 5 about point #(4 ,-6 ), how far will its centroid move?