If $f$ is an isometry of the plane and $L$ is a line, prove that $f(L)$ is a line.
I know that isometries preserves distance, so that is easy enough. I also know that two distinct point make up a line. Since we know that they share the same distance, I only have to prove that the image is a line. My question is how do I do that. Does proving that the image of $L$ is a line involve me showing that the distance of another point in $L$ is equidistant from the distinct point in $L$ and $f(L)$?
Here is what I would write down as a proof.
Let $L$ be the line containing $P$ and $Q$. Suppose there exists a point $S$ such that f(S) $\ne$ S. Since $f$ is an isometry we have $|PS| = |f(P)f(S)| = |P f(S)|.$ Similarly $|QS| = |f(Q)f(S)| = |Qf(S)|.$ Thus $P$ and $Q$ are equidistant from S and $f(S)$. Since $S$ $\ne$ $f(S)$, the set of points equidistant from $S$ and $f(S)$ is a line. This line contains $P$and $Q$ and thus equals L because $P \ne Q$. Therefore, every point on L is equidistant from $S$ and $f(S)$. And so $f(S)$ must lie on the line containing $P$ and $Q$ and the $f(L)$ is a line.
Is that close to being right?
This is a repost of my previous question. I deleted the last question.