# Isometries and Lines

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.

• For some basic information about writing math at this site see e.g. here, here, here and here. – Alex Provost Dec 15 '15 at 21:13
• When you copy-paste the output of a post written in MathJaX, you stumble into the funny issue that this post shows: all the parts that were originally written in LaTeX appear twice. – user228113 Dec 15 '15 at 21:14
• I will fix it, give me a minute. I apologize – Mark Dec 15 '15 at 21:15
• How do you know that $|f(P)f(S)|=|Pf(S)|$? – Aretino Dec 15 '15 at 21:20
• This is true of any invertible linear map, there is no need for it to be an isometry... – Morgan Rodgers Dec 15 '15 at 21:26

Use the property that $AB=AC+BC$ if and only if $A$, $B$, $C$ are on the same line and $C$ lies between $A$ and $B$.