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

Question

1452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-19T14:55:27

The distance is $=18.1$

The centroid of the triangle, $A=(-6,3)$ , $B=(3,-2)$ and $C=(5,7)$ is

$C_c=((-6+3+5)/3,(3-2+7)/3)$

$=(2/3,8/3)$

Let the point $D=(-2,-1)$

Let the centroid be $C_c'$ after the dilatation

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

$=5<2/3+2,8/3+1>$

$=<40/3,55/3>$

$C_c'=(40/3,55/3)+(-2,-1)$

$=(34/3,52/3)$

The distance between the centroids is

$d=sqrt((34/3-2/3)^2+(52/3-8/3)^2)$

$=sqrt((32/3)^2+(44/3)^2)$

$=1/3sqrt(32^2+44^2)$

$=1/3sqrt2960$

$=18.1$

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