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

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Answer 1 · 2017-06-12T12:51:30

The centroid will move by $=21.3$

The centroid of the triangle, $A=(3,5)$ , $B=(5,2)$ and $C=(9,4)$ is

$C_c=((3+5+9)/3,(5+2+4)/3)$

$=(17/3,11/3)$

Let the point $D=(1,9)$

Let the centroid be $C_c'$ after the dilatation

$vec(DC_c')=4vec(DC_c)$

$=4<17/3-1,11/3-9>$

$=<56/3,-64/3>$

$C_c'=(56/3,-64/3)+(1,9)$

$=(59/3,-37/3)$

The distance between the centroids is

$d=sqrt((59/3-17/3)^2+(-37/3-11/3)^2)$

$=sqrt((42/3)^2+(48/3)^2)$

$=sqrt(14^2+16^2)$

$=sqrt452$

$=21.3$

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