A triangle has corners at #(6 ,3 )#, #(4 ,-1 )#, and #(2 ,-9 )#. If the triangle is dilated by a factor of #5 # about point #(2 ,-4 ), how far will its centroid move?

1 Answer

It will move by distance#=4/3sqrt61=10.4137" "#units

Explanation:

We need to compute the old centroid and the new centroid first.

We compute the old centroid #(x_c, y_x)# first

#x_c=(x_1+x_2+x_3)/3=(6+2+4)/3=4#

#y_c=(y_1+y_2+y_3)/3=(3+(-9)+(-1))/3=-7/3#

Old centroid #(x_c, y_x)=(4, -7/3)#

Now we compute the new centroid #(x_c', y_c')#

The given problem is about dilated by a factor of 5, therefore the ratio of the distance of the new centroid from the reference point #(2, -4)# to the distance of the old centroid from the reference point is #5:1#

For #x_c'#

#(x_c'-2)/(x_c-2)=5/1#

#(x_c'-2)/(4-2)=5/1#

#x_c'=12#

For #y_c'#

#(y_c'--4)/(y_c--4)=5/1#

#(y_c'--4)/(-7/3--4)=5/1#

#y_c'=13/3#

New centroid #(x_c', y_c')=(12, 13/3)#

Distance #d# from old centroid to the new centroid

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

#d=sqrt((4-12)^2+(-7/3-13/3)^2)#

#d=sqrt(976/9)=4/3sqrt(61)=10.4137#

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