A triangle has corners at #(2 ,-5 )#, #(-8 ,4 )#, and #(1 ,-3 )#. If the triangle is dilated by a factor of #2/5 # about point #(-1 ,-2 ), how far will its centroid move?

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Answer 1 · 2018-02-26T05:12:02

Distance moved by centroid #color(green)(vec(GG’) = 0.9068#

Given : #A (2,-5), B(-8,4), C(1,-3)# Dilated about #D(-1,-2)# by a factor of #2/5#

To find how far the centroid has moved.

Centroid #G(x,y) = (2-8+1)/3, (-5+4-3)/3 => (-5/3, -4/3)#

#vec(A’D) = (2/5) vec(AD)#

#a’((x),(y)) - d((-1),(-2)) = (2/5)( a((2),(-5)) - d((-1),(-2)))#

#a’((x),(y)) = (2/5) ((2),(-5)) - (2/5)((-1),(-2)) + ((-1),(-2))#

#=> (2/5)((2),(-5)) + (3/5)((-1),(-2)) = ((4/5),(-2)) + ((-3/5),(-6/5))#

#a’((x),(y)) = ((1/5),(-16/5))#

Similarly, #b’((x),(y)) = (2/5)((-8),(4)) + (3/5)((-1),(-2)) = ((-16/5),(8/5)) + ((-3/5),(-6/5))#

#b’((x),(y)) = ((-19/5),(2/5))#

Similarly, #c’((x),(y)) = (2/5)((1),(-3)) + (3/5)((-1),(-2)) = ((2/5),(-3/5)) + ((-3/5),(-6/5))#

#c’((x),(y)) = ((-1/5),(-9/5))#

#G’(x,y) = (-3/5 -19/5 - 1/5)/3, (-16/5 + 2/5 - 9/5)/3) => (-23/15, —23/15)#

Distance moved by centroid

#vec(GG’) = sqrt((-5/3 + 23/15)^2 + (-4/3+23/15)^2) = 0.9068#