## Equidistant points from a pair of points

#### January 01, 2021

In the figure below, visually, it is not too hard to believe that the points equidistant from $u$ and $v$ are exactly the the points on the dashed line. A description of the dashed line seems to be “the line running perpendicularly through the midpoint of the segment between $u$ and $v$.” This turns out to be true, and perhaps is a standard exercise in Euclidean geometry. However, we can formulate this using linear algebra and obtain a result that generalizes to higher dimensions, yet the proof is still “strategized” by our visual intuition.

Theorem. Let $u$ and $v$ be two points in $\mathbb{R}^n$. Let $A \triangleq \{ w \in \mathbb{R}^n : \lVert w-u \rVert = \lVert w-v \rVert\}$ be the set of points equidistant from $u$ and $v$, and let $m \triangleq \dfrac{u+v}{2}$ be the midpoint of the segment between $u$ and $v$. Then, $A$ is equal to the the orthogonal complement of the line through $u$ and $v$ plus $m$; i.e.

$m + \text{span}(v-u)^\bot = A$

Proof. Let $m+w \in m + \text{span}(v-u)^\bot$, so that $\langle w, v-u \rangle = 0$. Note that $w$ and $m-u$ are orthogonal, as

\begin{aligned} \left\langle w, m-u \right\rangle &= \left\langle w, -\frac{u}{2}+\frac{v}{2} \right\rangle \\ &= \frac{1}{2} \langle w, v-u \rangle \\ &= 0. \end{aligned}

Similarly, $w$ and $m-v$ are orthogonal. As $m-u = -\left(m-v\right)$, we have $\left\lVert m-u \right\rVert = \left\lVert m-v \right\rVert$. Then, with the Pythagorean theorem,

\begin{aligned} \left\lVert \left(m+w\right)-u \right\rVert^2 &= \left\lVert m-u\right\rVert^2+ \lVert w \rVert^2 \\ &= \left\lVert m-v\right\rVert^2+ \lVert w \rVert^2 \\ &= \left\lVert \left(m+w\right)-v \right\rVert^2, \end{aligned}

so $m+w$ is equidistant from $u$ and $v$, and thus, $m + \text{span}(v-u)^\bot \subseteq A$.

Conversely, suppose $w \in A$, so that $\lVert w-u \rVert = \lVert v-u \rVert$. We can form a rhombus using $w$, $u$, and $v$ so that the diagonals of this rhombus are $(u-w)+(v-w)=u+v-2w$ and $(u-w)-(v-w)=v-u$. Note that for any $a$ and $b$, we have

\begin{aligned} \langle a+b, a-b \rangle &= \langle a, a \rangle - \langle a, b \rangle + \langle b, a \rangle - \langle b, b \rangle \\ &= \lVert a \rVert^2 - \lVert b \rVert^2. \end{aligned}

Therefore, we have $\langle u+v-2w, v-u \rangle = \lVert u-w \rVert^2 - \lVert v-w \rVert^2 = 0$, so $u+v-2w$ is orthogonal to $v-u$. Furthermore,

\begin{aligned} m-\frac{1}{2}(u+v-2w) &= \frac{u+v}{2} - \frac{u+v}{2} + w\\ &= w, \end{aligned}

so $w \in m+\text{span}(v-u)^\bot$, and thus, $m+\text{span}(v-u)^\bot = A$.