Point A is at #(5 ,2 )# and point B is at #(-6 ,-4 )#. Point A is rotated #(3pi)/2 # clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

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Answer 1 · 2018-02-10T11:54:35

Distance between A and B changed by #4.47# unit .

#A(5,2) and B (-6,-4)# . Clockwise rotation of point A is

#theta=(3pi)/2 :. # Counterclockwise rotation of point A is

#theta=(2pi)-(3pi)/2=pi/2#. New coordinates of #A(x',y')# can be

found by the fomula. #x'= xcos theta +ysin theta# and

#y'= y cos theta - x sin theta:.x'= 5*cos 90+ 2*sin90#

# :. x'= 5 *0 +2*1=2 ; y'= 2 * cos 90 - 5 * sin 90# or

#y'= 2 * 0- 5 * 1= -5# Therefore,new coordinates of A are

#(x',y') = (2,-5)# Distance between two points

#(x_1,y_1) and (x_2,y_2)# is #D= sqrt((x_1-x_2)^2+(y_1-y_2)^2)#.

Orginal distance between points #A(5,2) and B (-6,-4)# is

#D_o= sqrt((5+6)^2+(2+4)^2)=sqrt 157~~ 12.53# unit.

New distance between points #A(2,-5) and B (-6,-4)# is

#D_n= sqrt((2+6)^2+(-5+4)^2)=sqrt 65~~ 8.06# unit.

Distance between A and B changed by #12.53-8.06~~4.47# unit .

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