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)#