Point A is at #(-2 ,-4 )# and point B is at #(-3 ,6 )#. 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 · 2018-02-09T12:43:03

Distance between A and B changed by #5.93# unit.

#A(-2,-4) and B (-3,6)# . 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'= (-2)*cos 90+ (-4)*sin90 = -2 *0 -4*1#

#x' = -4 ; y'= -4 * cos 90+ 2 * sin 90 = -4 * 0+2 * 1= 2#

Therefore,new coordinates of A are #(x',y') = (-4,2)#

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(-2,-4) and B (-3,6)# is

#D_o= sqrt((-2+3)^2+(-4-6)^2)=sqrt 101~~ 10.05# unit.

New distance between points #A(-4,2) and B (-3,6)# is

#D_n= sqrt((-4+3)^2+(2-6)^2)=sqrt 17~~ 4.12# unit.

Distance between A and B changed by #10.05-4.12=5.93# unit .
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