A triangle has corners at #(-6 ,3 )#, #(3 ,-2 )#, and #(5 ,4 )#. If the triangle is dilated by a factor of #5 # about point #(-2 ,6 ), how far will its centroid move?

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Answer 1 · 2016-03-06T08:30:12

The centroid will move by about #d=4/3sqrt233=20.35245" "#units

We have a triangle with vertices or corners at the points #A(-6, 3) #and #B(3, -2)# and #C(5, 4)#.

Let #F(x_f, y_f)=F(-2, 6)" "#the fixed point

Compute the centroid #O(x_g, y_g)# of this triangle, we have

#x_g=(x_a+x_b+x_c)/3=(-6+3+5)/3=2/3#
#y_g=(y_a+y_b+y_c)/3=(3+(-2)+4)/3=5/3#

Centroid #O(x_g, y_g)=O(2/3, 5/3)#

Compute the centroid of the bigger triangle (scale factor =5)

Let #O'(x_g', y_g')=#the centroid of the bigger triangle

the working equation:

#(FO')/(FO)=5#

solve for #x_g'#:

#(x_g'--2)/(2/3--2)=5#

#(x_g'+2)=5*8/3#
#x_g'=40/3-2#
#x_g'=34/3#

solve for #y_g'#

#(y_g'-6)/(5/3-6)=5#

#y_g'=6+5(-13/3)=(18-65)/3#

#y_g'=-47/3#

Compute now the distance from centroid O(2/3, 5/3) to new centroid O'(34/3, -47/3).

#d=sqrt((x_g-x_g')^2+(y_g-y_g')^2)#

#d=sqrt((2/3-34/3')^2+(5/3--47/3)^2)#

#d=sqrt((-32/3)^2+(52/3)^2)#

#d=sqrt(((-4*8)/3)^2+((4*13)/3)^2)#

#d=4/3*sqrt(64+169)#

#d=4/3*sqrt(233)=20.35245#

God bless....I hope the explanation is useful..