 Solution

Let $\mathbf{u}=(2,1)/\sqrt{5}$ 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. ThereforeIts 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.