A line segment has endpoints at #(8 , 4)# and #(1 , 2)#. If the line segment is rotated about the origin by #(3pi)/2 #, translated vertically by #4#, and reflected about the x-axis, what will the line segment's new endpoints be?

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Feb 26, 2018

Answer:

Summarizing
transformed coordinates are
from
#x_11=8, y_11=4#
#x_21=1, y_21=2#
to
#x_14,y_14=4,4#
#x_24,y_24=2,-3#

Explanation:

#x_11=8, y_11=4#

#x_21=1, y_21=2#

#r_11=sqrt(x_11^2+y_11^2)=sqrt(8^2+4^2)=sqrt(64+16)=sqrt80#

#theta_11=tan^-1(y_11/x_11)=tan^-1(4/2)=26.565^@#

#r_21=sqrt(x_21^2+y_21^2)=sqrt(1^2+2^2)=sqrt(1+4)=sqrt5#

#theta_21=tan^-1(y_21/x_21)=tan^-1(2/1)=63.435^@^@#

Rotation is

#alpha=(3pi)/2=270^@#

#r_12=r_11#

#theta_12=theta_11+270^@#

#r_12=sqrt80#

#theta_12=26.565^@+270^@=296.565^@#

#r_22=r_21#

#theta_22=theta_21+270^@#

#r_22=sqrt5#

#theta_22=63.435^@+270^@=333.435^@#

#(x_12,y_12)=(r_12costheta_12,r_12sintheta_12)#
#=sqrt80cos296.565^@,sqrt80sin296.565^@=-=(4,-8)#

#x_12=4#
#y_12=-8#

#(x_22,y_22)=(r_22costheta_22,r_22sintheta_22)#
#=sqrt5cos333.435^@,sqrt5sin333.435^@=-=(2,-1)#
#x_22=2#
#y_22=-1#

translation is +4

#x_13=x_12#
#y_13=y_12+4#

#x_13=4#
#y_13=-8+4=-4#

#x_23=x_22#
#y_23=y_22+4#

#x_23=2#
#y_23=-1+4=3#

Reflection about x axis

#x_14=x_13#
#y_14=-y_12#

#x_14=4#
#y_14=4#

#x_24=x_23#
#y_23=-y_22#

#x_23=2#
#y_23=-3#

Summarizing

#x_11,y_11=8,4#
#x_21,y_21=1,2#
Rotation by #(3pi)/2#
#x_12,y_12=4,-8#
#x_22,y_22=2,-1#
translated bertically by 4
#x_13,y_13=4,-4#
#x_23,y_23=2,3#
reflected about x axis
#x_14,y_14=4,4#
#x_24,y_24=2,-3#

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