Isometries between two non parallel straight lines.

Question

for the past few days I've been studying Euclidean geometry and I was working on some problems concerning Euclidean transformations which are just isometries. The problem is the following: Let $l$ and $l'$ be two non parallel straight lines in $\mathbb{E}^2$. Show that there exist exactly 4 isomteries which project $l$ onto $l'$ and for which $l \cap l'$ is a fixed point. I know the rotation with center $l \cap l'$ and angle $\theta$ is one, where $\theta$ is the angle between $l$ and $l'$, as well as the rotation with center $l \cap l'$ with angle $\theta + \pi$ is another. However I can't seem to find two more and I have no clue to proof that there can't be more than 4. Does anyone have a tip in order to continue solving this problem? Any help would be greatly appreciated :))

Answer 1

Isometries of the Euclidean plane come in several types, including reflections. So there are indeed two more reflection isometries that take $l$ to $l'$, as follows: the lines $l$ and $l'$ determine two opposing pairs of angles; one reflection line bisects the angles in one of those opposing pairs; the other reflection line bisects the angles in the other opposing pair. For example, in the Cartesian coordinate model of Euclidean geometry, if $l$ is the line through the origin of slope $s>0$ and $l'$ is the line through the origin of slope $-s<0$ then reflection across the $x$-axis and reflection across the $y$-axis both take $l$ to $l'$.

To prove that these are the only four possibilities, let me introduce some notation. Let $P_1$, $P_2$ be the two points of $l$ at distance $1$ from the intersection point $O = l \cap l'$. Also let $Q_1$, $Q_2$ be the two points of $l'$ at distance $1$ from $l \cap l'$. For any isometry, we can choose $i \ne j \in \{1,2\}$ so that $P_1 \mapsto Q_i$ and $P_2 \mapsto Q_j$. There are exactly 4 choices of the ordered pair $(i,j)$. The SSS theorem implies the uniqueness of that isometry for each choice of $(i,j)$. For example, if $i=1$ and $j=2$ then the SSS theorem gives us the uniqueness of the isometry of the Euclidean plane that takes $\triangle OP_1P_2$ to $\triangle OQ_1Q_2$, with vertices mapped in the order shown.