A triangle has corners at #(-3 ,5 )#, #(7 ,6 )#, and #(1 ,-4 )#. If the triangle is dilated by a factor of #2/3 # about point #(-3 ,4 ), how far will its centroid move?

1 Answer
Mar 15, 2016

Answer:

Thus the new corners of the dilated triangle #A'B'C"# are
#A'(-3, 14/3), B'(11/3, 16/3), C'(-1/3, -4/3)#

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Explanation:

For each vertex we need to find the dilation of that point by 2/3 around (-3,4):
#A(-3,5), B(7, 6), C(1, -4)#
Dilate #B'=D_((-3,4), 2/3) *B' (7,6)=>[7-(-3), 6-4] *2/3 + [-3, 4]= (11/3, 16/3)#
#A'=D_((-3,4), 2/3)* A'(-3,5) =>[-3-(-3), 5-4] *2/3 + [-3, 4]= (-3, 14/3)#
#C'=D_((-3,4), 2/3) *C'(1,-4) =>[1-(-3), -4-4] *2/3 + [-3, 4]= (-1/3, -4/3)#

Thus the new corners of the dilated triangle #A'B'C"# are
#A'(-3, 14/3), B'(11/3, 16/3), C'(-1/3, -4/3)#