reflect a point over another point using matrix transformation We know that if we want to reflect any point over an origin, i.e. $ O\left(0, 0\right) $, we can use matrix transformation like this
$$ \left(\begin{matrix}x' \\ y'\end{matrix}\right) = \left(\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right)\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}-x \\ -y\end{matrix}\right). $$
But, what if we reflect any point over another point $ M\left(a, b\right) $ with $ a, b \ne 0 $?
 A: This would not generally be a linear transformation since $(0,0)$ would not map to itself via a reflection over a non-origin point. So you will not be able to do this with a single matrix multiplication.
I would suggest translating the coordinate system so that the reflection point is at the new origin; then reflect; and then translate back.
For example, to translate $(x,y)$ over $(5,7)$, I would do
$\left(\begin{array}{c} x' \\ y' \end{array}\right)=\left(\begin{array}{c} x-5 \\ y-7 \end{array}\right)\left(\begin{array}{cc} -1&0 \\0&-1 \end{array}\right)+ \left(\begin{array}{c} 5 \\ 7 \end{array}\right)$
A: Another way to describe the point reflection across $\,M\left(a, b\right)\,$  which maps $\,P\left(x,y\right)\,$ to $\,P'(x',y')\,$ is that the midpoint of $\,PP'\,$ is always $\,M\,$.

*

*In matrix notation: $$\displaystyle \quad\quad\dfrac{1}{2}\,\Bigg(\left(\begin{matrix}x \\ y\end{matrix}\right) + \left(\begin{matrix}x' \\ y'\end{matrix}\right)\Bigg) =\left(\begin{matrix}a \\ b\end{matrix}\right)
\\ \iff\quad \left(\begin{matrix}x' \\ y'\end{matrix}\right) = \left(\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right)\left(\begin{matrix}x \\ y\end{matrix}\right) + \left(\begin{matrix}2a \\ 2b\end{matrix}\right) = \left(\begin{matrix}2a - x \\ 2b - x\end{matrix}\right) \text{.}\,$$


*In vector notation: $\quad \dfrac{1}{2}\,\left(\overrightarrow{OP} + \overrightarrow{OP'}\right) = \overrightarrow{OM}\;\iff\; \overrightarrow{OP'} = 2\, \overrightarrow{OM} - \overrightarrow{OP}\,$.


*In complex notation ($\,p = x + i y\,$, $\,m = a + ib\,$): $\quad \dfrac{1}{2}\left(p+p'\right) = m \;\iff\; p' = 2m - p\,$.
