A triangle has corners at #(6 ,2 )#, #(5 ,-8 )#, and #(-5 ,3 )#. If the triangle is dilated by a factor of #5 # about point #(7 ,-2 ), how far will its centroid move?

1 Answer
Douglas K.
Oct 10, 2016

The distance moved is #4sqrt(26)#

Explanation:

Let's begin by computing the current centroid, point #O_1#,

#O_(1x) = (6 + 5 - 5)/3 = 2#

#O_(1y) = (2 - 8 + 3)/3 = -1#

Point #O_1 = (2, -1)#

  1. Scale point #(6, 2)#:

Compute the vector from point #(7, -2)# to point #(6, 2)#:

#(6 - 7)hati + (2 - -2)hatj = -hati + 4hatj#

Scale by 5:

#-5hati + 20hatj#

Find the new end point:

#(-5+ 7, 20 + -2) = (2, 18)#

  1. Scale point #(5, -8)#:

Compute the vector from point #(7, -2)# to point #(5, -8)#:

#(5 - 7)hati + (-8 - -2)hatj = -2hati -6hatj#

Scale by 5:

#-10hati - 30hatj#

Find the new end point:

#(-10+ 7, -30 + -2) = (-3, -32)#

  1. Scale point #(-5, 3)#:

Compute the vector from point #(7, -2)# to point #(-5, 3)#:

#(-5 - 7)hati + (3 - -2)hatj = -12hati + 5hatj#

Scale by 5:

#-60hati + 25hatj#

Find the new end point:

#(-60+ 7, 25 + -2) = (-53, 23)#

Our scaled triangle has the vertices, #(2, 18)#, #(-3, -32)#, and #(-53, 23)#. The new centroid, #O_2#, has coordinates:

#O_(2x) = (2 - 3 - 53)/3 = -18#

#O_(2y) = (18 - 32 + 23)/3 = 3#

#O_2 = (-18, 3)#

Compute the distance, d, between #O_1# and #O_2#:

#d = sqrt((-18 - 2)^2 + (3 - -1)^2)#

#d = sqrt((-20)^2 + 4^2)#

#d = sqrt((-20)^2 + 4^2)#

#d = sqrt(416)#

#d = 4sqrt(26)#

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