= Solution
Let $\mathbf u=(2,1)/\sqrt5$ be a unit <vector> along the fixed line. The <orthogonal projection> parallel to the line is $\mathbf u\mathbf u^T$; a <reflection> keeps that component and reverses its orthogonal complement. Therefore
$$
\boxed{A=2\mathbf u\mathbf u^T-I=\frac15\begin{pmatrix}3&4\\4&-3\end{pmatrix},\qquad \det A=-1.}
$$
Its <eigenvalues> are $+1$ along $(2,1)$ and $-1$ along $(-1,2)$. It preserves lengths and unsigned areas but reverses orientation. For example, $(1,0)$ maps to $(3/5,4/5)$, with their midpoint on the reflection line and their separation perpendicular to it; the right panel of the figure in <1> illustrates this.
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