I was looking at the following question.
Given two non-parallel lines $l$ and $l'$ in $\mathbb{E}^2$. Show that there exist exactly 4 isometries that map $l$ to $l'$ and for which the point $x_0$ with $l \cap l' = \{x_0\}$ is a fixed point.
My attempt:
Using the classification theorem for isometries in $\mathbb{E}^2$, it would suffice to show that there are exactly 0 translations, exactly 2 rotations and 2 glide reflections which meet the conditions.
Suppose there is a translation $t_b$ that satisfies the conditions. Then because $x_0$ is fixed, $t_b(x_0) = x_0 + b = x_0$ which implies $b = 0$. But then $t_b$ is the identical map which would imply $l = l'$. This contradicts the assumption that the lines are not parallel.
Now suppose that $F$ is a rotation that satisfies the conditions. Start by noticing that then $x_0$ is necessarily the center of the rotation because the only fixed point of a rotation is its center. Now my idea was to prove that the only angles of the rotation that satisfy the conditions would be the two angles $\theta_1$ and $\theta_2$ with $\cos(\theta_{1,2}) = \frac{r \cdot s}{\lVert r \rVert \lVert s \rVert}$ where $r$ and $s$ are the directions of the lines $l$ and $l'$ respectively. Visually this would be logical. I seem to be stuck however at this point.
Showing that there are exactly two glide reflections satisfying the conditions visually makes sense to me but I also seem to be stuck showing that this is the case.
A very similar question was also asked here Formally showing that there exist exactly four isometries of $\mathbb{E}^2$ that map two intersecting lines, but the answer didn't seem correct to me. Also here the lines were assumed to be intersecting which doesn't exclude the case where $l = l'$.
Thank you for your time.