Relation between rotating a point with respect to a point, with translation

Question

A little bit ago in class, we learned about how to rotate a point with respect to a point.

If we're talking about middle school math, usually we rotate with respect to the origin $(0, 0)$.

The general formula for rotating a point $(x, y)$ A degrees with respect to the origin is $$x' = x\cos(A) - y\sin(A)$$ $$y' = x\sin(A) + y\cos(A)$$

But, if we rotate with respect to (m, n), is $$x' = (x-m)\cos(A) - (y-n)\sin(A) + m$$ $$y' = (x-m)\sin(A) + (y-n)\cos(A) + n$$ It's better to use matrices here, but I can't get it to work. (Someone please edit it)

And that got me thinking, is rotating with respect to $(m, n)$ the same as rotating with respect to $(0, 0)$ and then translating the result by $(m, n)$?

Since rotating a point is only dependent on the angle we rotate with, then by translating the result does not affect "how much the point rotated." And if we rotate a point with respect to a point $(m, n)$, we can just assume $(m, n)$ as the origin and do rotation wrt the assumed origin like normal. Then considering where $(m, n)$ is, we possibly can find where the point lands wrt the original origin $(0,0)$ (I say possibly because I'm unsure.)

Its hard to visualize this so I decided to ask: is rotating with respect to $(m, n)$ the same as rotating with respect to $(0, 0)$ and then translating the result by $(m, n)$?

Answer 1

The representation using matrices would be $R(x,y)= \left ( \begin{array} {ccc} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha)\end{array}\right) \left( \begin{array}{ccc} x \\ y \\ \end{array}\right) $, where $\alpha$ is the angle of rotation.
In general, this is not true. Take for instance $(1,0)$ and a rotation of $\alpha = \pi/2$ (with respect to the origin). This is now $(0,1)$. If we compose it with a traslation with vector $v = (1,1),$ the final point would be $(1,2)$. Let's do it on the other way. If one first applies to $(1,0)$ the traslation, one obtains $(2,3)$. The rotation of $\alpha = \pi/2$ degrees now gives another point which is not $(1,2)$ (you can calculate the point using the matrix expression).

Answer 2

Rotating about $(m, n)$ is not the same as rotating about $(0,0)$ and translating by $(m, n)$. To see this note that former leaves $(m, n)$ unchanged while later will always move it to a different point. What's true however is that rotating about $(m, n)$ is same as first translating $(m, n)$ to $(0,0)$, then rotating about $(0,0)$ and then translating everything back by $(m, n)$. So first you take $(x, y)$ to $(x-m, y -n)$. Then rotate it about origin by angle $\theta$ which gives you

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}x -m \\ y-n\end{bmatrix}$

And finally translating everything back by $(m, n)$ gives you

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}x -m \\ y-n\end{bmatrix} + \begin{bmatrix} m \\ n\end{bmatrix}$