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

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Answer 1 · 2017-06-10T06:39:21

The distance is #=4.62#

Let the centroid of the triangle be #C#

Then,

#C=((9-2+3)/3,(-5-1-4)/3)#

#=(10/3,-10/3)#

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

The point #D=(1,4)#

So,

#vec(DC')=2/5vec(DC)#

#((x'-1),(y'-4))=2/5((10/3-1),(-10/3-4))#

#x'-1=2/5(10/3-1)=14/15#

#=>#, #x'=14/15+1=29/15#

#y'-4=2/5(-10/3-4)=-44/15#

#y'=-44/15+4=16/15#

Therefore,

The new centroid is #C'=(29/15,16/15)#

The distance

#CC'=sqrt((10/3-29/15)^2+(-10/3-16/15)^2)#

#=sqrt((21/15)^2+(66/15)^2)#

#=1/15sqrt(21^2+66^2)#

#=4.62#