Which lines lie equidistant from two points?

Question

Note that it's not the locus of points equidistant from the two points, but the collection of lines that are equidistant from two points.

I considered the two points to be the diametrically opposite ends of a circle;

It's obvious that all the lines that pass through the centre of this circle will be equidistant from the two points. All horizontal lines satisy the condition, too.

Are there any more lines that are equidistant from any two arbitrary points?

Answer 1

Let point $A$ be $(-k,0)$ and point $B$ be $(k,0)$.
Let $y=mx+c$ be the equation of line equidistant from $A$ and $B$*. We have,
$$\frac{|-km+c|}{\sqrt{1+m^2}}=\frac{|km+c|}{\sqrt{1+m^2}}$$ $$⇒|-km+c|=|km+c| $$ $$⇒-km+c=±(km+c) $$ If$-km+c=-(km+c) $ then $c=0$
$c=0$ means lines passing through origin

If$-km+c=km+c $ then $km=0 $ $ ⇒m=0$
$m=0$ means lines parallel to x-axis

So lines passing through origin and lines parallel to x-axis are the only lines that satisfy the condition.

• Lines parallel to the Y-axis, as specified by @TonyK, cannot be expressed in this general form, and must be separately considered. Obviously, none of them are equidistant from the points, and thus can all be ruled out (except the Y-axis itself, which is obtained from the first part).

Answer 2

Let $\ell$ be such a line, and let $P$ and $Q$ be the points on $\ell$ nearest to $A$ and $B$ respectively. $P$ coincides with $A$ if and only if $Q$ coincides with $B$, in which case $\ell$ is the $x$-axis. Otherwise, $AP$ and $BQ$ are perpendicular to $\ell$, and we get two cases, depending on whether $A$ and $B$ are on the same side of $\ell$ or not:

1. If they are on the same side of $\ell$, then line segments $APQB$ form three sides of a rectangle, so $\ell$ is parallel to the $x$-axis.
2. Otherwise we get:

In this case the line segments $APQB$ are rotationally symmetric, so $PQ$ passes through the centre $O$.

Answer 3

Let the points be $(\pm1,0)$. We have

$$d=|x\cos\theta+y\sin\theta+p|$$ and we express the equality of the distances:

$$|\cos\theta+p|=|-\cos\theta+p|.$$

The solutions are given by $$\cos\theta=0,$$ or $$p=0.$$

Hence we have the possible equations

$$\pm y+p=0,$$ a family of horizontal lines and

$$x\cos\theta+y\sin\theta,$$ a family of lines through the origin.