How do I find transformation rotation?

[footprints]

1)Two operations, R and T, are defined as follows:
R is a clockwise rotation through 90$$\circ$$ about (0,2) and T is a translation which maps the point (x,y) onto the point (x+3, y+1).
If P is the point (2,3), find the coordinates of R(P) and TR(P)

How do I find rotation? I'm just stuck with that part.

 

[ceptimus]

Draw a few points on a graph. Rotate them around the point (0, 2) by 90 degrees and see where they end up.

I think you'll get something like:

(Xr, Yr) = (Y-2, 2-X)

But check it, as I might be wrong.

 

[HallsofIvy]

What exactly do you mean by "find the transformation". A standard technique for a rotation around (x₀, y₀) is: translate (add or subtract) to move (x₀, y₀) to (0,0), then rotate around (0,0), then translate back again.

In the case of "clockwise rotation about (0,2) by 90 degrees", first move the point (x,y) to (x, y- 2) (so that (0,2) would be moved to (0,0). Rotating clockwise 90 degrees about (0,0) moves the point (x,y) to (-y, x) so (x, y-2) would be moved to
(2-y, x). Now translate back again: (2-y, x) moves to (2-y, x+2).
That is, a "clockwise rotation about (0,2) by 90 degrees" moves the point (x,y) to
(2-y, x+2). Notice that (0,2) itself would be moved to (2-2, 0+2)= (0,2) again. That's obviously correct, the point about which we rotate should not change. On the other hand, the point (1,2) is 1 unit directly to the right of (0,2) and a "clockwise rotation about (0,2) by 90 degrees" should move it to a point 1 unit above (0,2)- that is, (0,3). Yes, with x= 1, y= 2, (2-y,x+2) becomes (2-2,1+2)= (0,3).

In particular, if P= (2, 3) then R(P)= (2-3,2+2)= (-1, 4) and TR(P)= (-1+3, 4+1)= (2,5).

 

[ceptimus]

Doesn't HallsofIvy's translation rotate anti-clockwise, given the conventional orientation of the x and y axes?

 

[footprints]

What exactly do you mean by "find the transformation". A standard technique for a rotation around (x₀, y₀) is: translate (add or subtract) to move (x₀, y₀) to (0,0), then rotate around (0,0), then translate back again.

I'm not entirely sure what you mean by find the transformation because I don't remember typing it anywhere. If you are referring to my title, then that is what the subject is about.

In the case of "clockwise rotation about (0,2) by 90 degrees", first move the point (x,y) to (x, y- 2) (so that (0,2) would be moved to (0,0). Rotating clockwise 90 degrees about (0,0) moves the point (x,y) to (-y, x) so (x, y-2) would be moved to
(2-y, x). Now translate back again: (2-y, x) moves to (2-y, x+2).
That is, a "clockwise rotation about (0,2) by 90 degrees" moves the point (x,y) to
(2-y, x+2). Notice that (0,2) itself would be moved to (2-2, 0+2)= (0,2) again. That's obviously correct, the point about which we rotate should not change. On the other hand, the point (1,2) is 1 unit directly to the right of (0,2) and a "clockwise rotation about (0,2) by 90 degrees" should move it to a point 1 unit above (0,2)- that is, (0,3). Yes, with x= 1, y= 2, (2-y,x+2) becomes (2-2,1+2)= (0,3).

That is exactly what I'm looking for (not ceptimus's first reply. Not that I'm ungrateful or anything), except that the books answers doesn't match yours. As ceptimus said, I think you gave me the translation for anti-clockwise rotation. If I'm not wrong, a clockwise rotation about (0,0) by 90 degrees should be $$(x,y) \rightarrow (y,-x)$$ right? My problem is that I'm not sure where the numbers go.

 

[HallsofIvy]

Yes, I was doing it counter-clockwise instead of clockwise! Rotating clockwise about (0,0) takes (x, y) into (y, -x) so: (x,y) translates to (x, y-2), then rotates to (y-2,-x) then translates back to (y-2,2-x) which was what you said originally wasn't it?