A triangle has corners at $(8 ,2 )$ , $(4 ,9 )$ , and $(7 ,3 )$ . If the triangle is dilated by a factor of $5$ about point #(1 ,-6 ), how far will its centroid move?

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Answer 1 · 2017-03-17T16:05:29

THe distance between the centroids is $=47.7$

Let $ABC$ be the triangle

$A=(8.2)$

$B=(4,9)$

$C=(7,3)$

The centroid of triangle $ABC$ is

$C_c=((8+4+7)/3,(2+9+3)/3)=(19/3,14/3)$

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

The center of dilatation is $D=(1,-6)$

$vec(DA')=5vec(DA)=5*<7,8> = <35,40>$

$A'=(35+1,40-6)=(36,34)$

$vec(DB')=5vec(DB)=5*<3,15> = <15,75>$

$B'=(15+1,75-6)=(16,69)$

$vec(DC')=5vec(Dc)=5*<6,9> = <30,45>$

$C'=(30+1,45-6)=(31,39)$

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

$C_c'=((36+16+31)/3,(34+69+39)/3)=(83/3,142/3)$

The distance between the 2 centroids is

$C_cC_c'=sqrt((83/3-19/3)^2+(142/3-14/3)^2)$

$=1/3sqrt(64^2+128^2)=143.1/3=47.7$

A triangle has corners at (8 ,2 )(8 ,2 ), (4 ,9 )(4 ,9 ), and (7 ,3 )(7 ,3 ). If the triangle is dilated by a factor of 5 5 about point #(1 ,-6 ), how far will its centroid move?