A line segment goes from #(5 ,2 )# to #(4 ,2 )#. The line segment is reflected across #x=-3#, reflected across #y=-5#, and then dilated about #(2 ,0 )# by a factor of #2#. How far are the new endpoints from the origin?

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Answer 1 · 2018-03-05T06:32:29

Distance of new points from origin

#vec(OA”) = 33.94#

#vec(OB”) = 32.56#

Given points #A(5,2) , B (4,2)#

Reflected across #x = -3#, #y = -5#

![Given points #A(5,2) , B (4,2)#

Reflected across #x = -3#, #y = -5#

#A (x,y) -> A’(x, y) = (5,2) -> ((2x’ - 5), (2y’ - 2)) => (-11, -12)#

#A’(x,y) => ((-6 - 5), (-10 - 2)) => (-11, -12)#

#B (x,y) - > B’ (x,y) = (4 , 2) -> ((2x’ - 4), (2y’ - 2)) #

#B’(x,y) => ((-6 - 4), (-10 - 2)) => (-10, -12)#

New coordinates after reflection are #A’(-11, -12), B’(-10, -12)#

Now we we have to find A”, B” after rotation about point C (2,0) with a dilation factor of 2.

#A”(x, y) -> 2 * A’(x,y) - C (x,y)#

#A”((x),(y)) = 2 * ((-11), (-12)) - ((2), (0)) = ((-24),(-24))#

Similarly, #B”((x),(y)) = 2 * ((-10), (-12)) - ((2), (0)) = ((-22),(-24))#

New endpoints are #A”(-24, -24), B”(-22, -24)#

Distance of new points from origin

#vec(OA”) = sqrt(-24^2 + -24^2) = 33.94#

#vec(OB”) = sqrt(-22^2 + -24^2) = 32.56#]

#A (x,y) -> A’(x, y) = (5,2) -> ((2x’ - 5), (2y’ - 2) => (-11, -12)#

#A’(x,y) => ((-6 - 5), (-10 - 2) => (-11, -12)#

#B (x,y) - > B’ (x,y) = (4 , 2) -> ((2x’ - 4), (2y’ - 2) #

#B’(x,y) => ((-6 - 4), (-10 - 2) => (-10, -12)#

New coordinates after reflection are #A’(-11, -12), B’(-10, -12)#

Now we we have to find A”, B” after rotation about point C (2,0) with a dilation factor of 2.

#A”(x, y) -> 2 * A’(x,y) - C (x,y)#

#A”((x),(y)) = 2 * ((-11), (-12)) - ((2), (0)) = ((-24),(-24))#

Similarly, #B”((x),(y)) = 2 * ((-10), (-12)) - ((2), (0)) = ((-22),(-24))#

New endpoints are #A”(-24, -24), B”(-22, -24)#

Distance of new points from origin

#vec(OA”) = sqrt(-24^2 + -24^2) = 33.94#

#vec(OB”) = sqrt(-22^2 + -24^2) = 32.56#