We consider the linear transformation in a plane consisting of the symmetry with regards to the line $L$: $ax + by = 0$
Each point $M=(x, y)$ of the plane is sent to its symmetric point across line (L).
$M=(x,y) \to M'=(x',y')$ such that line $MM'$ is orthogonal to $L$ and the distances $MN$ and $M'N$ are equal where $N$ is the intersect between line $MM'$ and line $L$
Find the $2\times 2$ matrix $S$ representing this linear transformation.
(Hint: first find a set of orthonormal eigenvectors of the transformation and their eigenvalues. Then write matrix $S$ as the product of $PDP'$ where $D$ is a diagonal matrix and $P'$ is the transpose of $P$.)
Use the Hint !
The points of the line $L$ are fixed points so, obviously, any vector on the line $L$ is an eigenvector of the eigenvalue $1$. A normal such vector is $$ v_{(1)}=\frac{1}{\sqrt{a^2+b^2}}\begin{bmatrix} b\\-a \end{bmatrix} $$
Also, the vectors of the line orthogonal to $L$ (that has equation $bx-ay=0$) are eigenvectors of the eigenvalue $-1$ (can you see why?) and a normal vector of this kind is $$ v_{(-1)}=\frac{1}{\sqrt{a^2+b^2}}\begin{bmatrix} a\\b \end{bmatrix} $$
Can you do from this?