Elementary rotations I

The rotation of a point $(a,b)$ about the origin in the coordinate plane is equivalent to the transformation defined by $(a,b) \mapsto (-b,a)$ . However, we want to rotate the point $Q(a,b)$ about $P(x,y)$ . Since we are working in $\mathbb{R}^2$ , one simple way to start doing this would be to translate every point in the plane by $T_{-x,-y}$ , such that $(x,y) \mapsto (0,0)$ , and $(a,b) \mapsto (a-x,b-y)$ . Now we are rotating the point $(a-x,b-y)$ about the origin. Clearly, the image of this point under a counterclockwise $90^{\circ}$ rotation about the origin is $(y-b,a-x)$ . But this is after both points were translated in the plane so that one of them went to the origin. In other words, we find the correct coordinates for the image of the rotation by 'undoing' the initial translation we did in the first place with the inverse translation $T^{-1}_{x,y}$ , such that $(0,0) \mapsto (x,y)$ , and $(y-b,a-x) \mapsto (x+y-b,a+y-x)$ . Since the rotation is not affected by translations, we see that the point $P'(x+y-b,a+y-x)$ is the image of $(a,b)$ under a counterclockwise $90^{\circ}$ rotation about $(x,y)$ .

Now, we are told that the coordinates of $P'$ are $(x+y,y-x)$ . Then clearly we must have the following: $x+y-b=x+y \implies b=0 \\ a+y-x=y-x \implies a=0$ Therefore, we see that the coordinates of $Q(a,b)$ are $(0,0)$ , so $ab=0$ .