I always have trouble visualizing these as I am trying to figure out the single transformation that is equivalent to the below compositions.
A 90 degree rotation about the origin, followed by a 90 degree rotation about the point (0,6).
R (0,6),90° ∘ R (0,0),90°- My idea is to break the two rotations into two reflections each (so 4 reflections). I am assuming two of the reflections would cancel and be left with two reflections, which would be a single rotation, but I am stuck on what reflections they break down to.
I think I am on the right track, but who knows. Any help would be very much appreciated.
I claim that the first composition is the $180^\circ$ rotation around the point $(3,3)$.
After the first rotation (the rotation around the origin), we have $(x,y) \mapsto (-y,x)$.
Now, we want to rotate the point $(-y,x)$ $90^\circ$ around (0,6). The vector pointing from $(0,6)$ to $(-y,x)$ is $$ \vec{v} = (-y,x-6) $$ To rotate this vector $90^\circ$, we do the same transformation as before (I mean $(a,b) \to (-b,a)$), so the rotated vector is $$ \vec{w} = (6-x,-y) $$ We should be thinking of this vector based at the point $(0,6)$, so its endpoint is at $$ (0,6) + \vec{w} = (6-x,6-y) $$
So in conclusion, the composition of these two rotations sends any point $(x,y)$ to the point $(6-x,6-y)$. Notice that the only point which is fixed is $(3,3)$. You can see this just by solving the equations $6-x=x$ and $6-y=y$. Since there is only one fixed point, this can't be a reflection (reflections fix all the points on a line), so it must be some rotation around $(3,3)$. You can do a little more work to see that it's a 180-degree rotation.