Point A is at #(-8 ,-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?

The rotation of #pi/2# clockwise about the origin transforms the point #A# into #A'#

The coordinates of #A'# are

#((0,1),(-1,0))*((-8),(-7))=((-7),(8))#

Distance #AB# is

#=sqrt((-3+8)^2+(-4+7)^2)#

#=sqrt(34)#

Distance #A'B# is

#=sqrt((-3+7)^2+(-4-8)^2)#

#=sqrt160#

The distance has changed by

#=sqrt160-sqrt34#

#=6.82#

When point A is rotated Π/2 clockwise, it moves up and crosses X axis.

For, the original angle created by point A at origin with X axis was less than Π/2.

Now, the distance OA (where, O is origin) is a constant.

Let us designate A' as new position of point A, after rotating Π/2 clockwise.

So we get an isosceles triangle AA'O, where angle AOA' is Π/2.

An isosceles triangle having Π/2 as the angle at vertex, will cause base angles as Π/4.

Now base of the triangle AA'O cuts X axis. Let us designate this point as O'

In triangles A'O'O and AO'O, OO' is a common arm, ∠O'A'O = ∠O'AO =Π/4; for triangle A'OA is an isosceles triangle.

So, triangle A'O'O and triangle AO'O are equal in all respects.

Hence O'A'=O'A=7 units; for coordinate of A was (-8, -7).

So coordinate of A' will be (-8, +7).

Now, distance between two points in Cartesian Coordinates is
#((X1-X2)^2+(Y1-Y2)^2)^(1/2)#, where X1, Y1 and X2,Y2 are the coordinates of the two points.

Original distance was #((-7+3)^2+(-8+4)^2)^.5 #=#32^.5#

New distance is #((-7+3)^2+(+8+4)^2)^.5 #=#160^.5#

Therefore, we can say that distance between points A and B has changed.