A line segment has endpoints at #(3 ,5 )# and #(6 ,1 )#. The line segment is dilated by a factor of #4 # around #(6 ,2 )#. What are the new endpoints and length of the line segment?

Answer 1

Theodore Abad

New points #A’ ((-6),(14))#, #B’ ((6),(-2))#

Length of the line segment after dilation #vec(A’B’) = 18.44#

#A (3, 5), B ( 6, 1)#. Dilated around #C (6,2 # by factor 4.

#vec (A’C) = 4 * vec(AC)#

#A’ - c= 4 * (a - c)#

#A’ = 4a - 3c#

#A’((x),(y)) = 4 ((3),(5)) - 3((6),(2)) = ((12),(20)) - ((18),(6)) = ((-6),(14))#

#A’ ((-6),(14))#

Similarly,

#vec(B’C) = 4 vec(AB)#

#B’((x),(y)) = 4((6),(1)) - 3((6),(2)) = ((24),(4)) - ((18),(6)) = ((6),(-2))#

#B’ ((6),(-2))#

Using distance formula, #vec(AB) = sqrt((3-6)^2 + (5-1)^2) = 5#

#vec(A’B’) = sqrt((-6-6)^2 + (14+2)^2) ~~ 18.44#

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Answer 2

Theodore Abad

The new endpoints of the line segment after dilation by a factor of 4 around (6, 2) are (15, 14) and (18, 10). The length of the line segment remains the same after dilation, which is ( \sqrt{(18 - 15)^2 + (10 - 14)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 ).

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