Question

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?

Answer 1

The new coordinates are `=(-7,8)` and the distance has changed by `=6.82`

Explanation:

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`

Answer 2

New coordinates of point A is (-8,7) and distance between points A and B has changed.

Explanation:

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.