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

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Answer 1 · 2017-12-21T14:35:14

The centroid will move by #=8.6u#

The corners of the triangle are #A=(9,5)#, #B=(4,-9)# and #C=(2,-4)#

The centroid of triangle #ABC# is

#C=((9+4+2)/3, (5-9-4)/3)=(15/3,-8/3)#

Let the the new centroid be #C'=(x,y)# after dilatation.

Let the fixed point be #D=(1,4)#

Let the coefficient of dilatation be #k=7/4#

Therefore, in vector notation

#vec(DC')=kvec(DC)#

#((x-1),(y-4))=7/4((15/3-1),(-8/3-4))#

#x-1=7/4(*4)#, #=>#, #x=8#

#y-4=7/4*(-20/3)#, #=>#, #y=-35/3+4=-23/3#

The new coordinates of the centroid are #C'=(8,-23/3)#

The centroid will move by

#=sqrt((8-1)^2+(-23/3+8/3)^2)#

#=sqrt(49+25)#

#=sqrt(74)#

#=8.6u#