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

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Answer 1 · 2016-03-31T06:53:23

The displacement from old centroid to the new centroid is
#d=sqrt596#
#d=24.413" "#units

From the given, let #P_1(x_1, y_1)=(3, 1)# and #P_2(x_2, y_2)=(5, 2)# and #P_3(x_3, y_3)=(9, 4)# and #R(1, 9)#

Compute the centroid #C_1(x_c, y_c)# of the triangle first

#x_c=(x_1+x_2+x_3)/3=(3+5+9)/3=17/3#
#y_c=(y_1+y_2+y_3)/3=(1+2+4)/3=7/3#

The centroid #C_1(x_c, y_c)=(17/3, 7/3)#

Determine the new centroid #C_2(x_c', y_c')#

The ratio:
#(C_2 R)/(C_1 R)=4/1#
Find #x_c'#
#(x_c'-1)/(17/3-1)=4/1#

#x_c'=1+4(17/3-1)#

#x_c'=1+56/3#

#x_c'=59/3#

Find #y_c'#

#(y_c'-9)/(7/3-9)=4/1#

#y_c'=9+4(7/3-9)#

#y_c'=9+(4(7-27))/3#
#y_c'=9-80/3#
#y_c'=(27-80)/3#

#y_c'=-53/3#

The New Centroid is at #C_2(59/3, -53/3)=(19.67, -17.67)#

Let us determine the displacement #d#

#d=sqrt((x_c-x_c')^2+(y_c-y_c')^2)#

#d=sqrt((59/3-17/3)^2+(-53/3-7/3)^2)#

#d=sqrt(14^2+(-20)^2)#
#d=sqrt596#
#d=24.413" "#units

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