## Question

###### Point A is at #(3 ,7 )# and point B is at #(3 ,-4 )#. Point A is rotated #pi/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?

## Answers

New coordinates of point A is#(-7,3)# and Distance between A and B changed by#1.1# unit.## Explanation:

#A(3,7) and B (3,-4)# . Clockwise rotation of point A is

#alpha=pi/2 :. # Counterclockwise rotation of point A is

#theta=2pi-alpha=2pi-pi/2=(3pi)/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'= 3*cos((3pi)/2)+7*sin((3pi)/2) = 0+(-7)=-7#

#x'= -7; y'= 7* cos((3pi)/2)- 3 * sin((3pi)/2) #

#= 7 * 0-3 *(-1)=3:. y'=3 :. (x',y') = (-7,3)# 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 betweenpoints

#A(3,7) and B (3,-4)# is

#D_o= sqrt((3-3)^2+(7+4)^2)=sqrt 121= 11.0# unit.New distance between points

#A(-7,3) and B (3,-4)# is

#D_n= sqrt((-7-3)^2+(3+4)^2)=sqrt 149~~ 12.2# unitDistance between A and B changed by

#12.2-11.1=1.1# unit .[Ans]