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 uu and vv 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 uu and vv.” 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 uu and vv be two points in Rn\mathbb{R}^n. Let A{wRn:wu=wv}A \triangleq \{ w \in \mathbb{R}^n : \lVert w-u \rVert = \lVert w-v \rVert\} be the set of points equidistant from uu and vv, and let mu+v2m \triangleq \dfrac{u+v}{2} be the midpoint of the segment between uu and vv. Then, AA is equal to the the orthogonal complement of the line through uu and vv plus mm; i.e.

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

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

w,mu=w,u2+v2=12w,vu=0.\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, ww and mvm-v are orthogonal. As mu=(mv)m-u = -\left(m-v\right), we have mu=mv\left\lVert m-u \right\rVert = \left\lVert m-v \right\rVert. Then, with the Pythagorean theorem,

(m+w)u2=mu2+w2=mv2+w2=(m+w)v2,\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+wm+w is equidistant from uu and vv, and thus, m+span(vu)Am + \text{span}(v-u)^\bot \subseteq A.

Decomposition of the distance from uu [or vv] to our given vector, m+wm+w, into orthogonal parts.

Conversely, suppose wAw \in A, so that wu=vu\lVert w-u \rVert = \lVert v-u \rVert. We can form a rhombus using ww, uu, and vv so that the diagonals of this rhombus are (uw)+(vw)=u+v2w(u-w)+(v-w)=u+v-2w and (uw)(vw)=vu(u-w)-(v-w)=v-u. Note that for any aa and bb, we have

a+b,ab=a,aa,b+b,ab,b=a2b2.\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 u+v2w,vu=uw2vw2=0\langle u+v-2w, v-u \rangle = \lVert u-w \rVert^2 - \lVert v-w \rVert^2 = 0, so u+v2wu+v-2w is orthogonal to vuv-u. Furthermore,

m12(u+v2w)=u+v2u+v2+w=w,\begin{aligned} m-\frac{1}{2}(u+v-2w) &= \frac{u+v}{2} - \frac{u+v}{2} + w\\ &= w, \end{aligned}

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

Forming a rhombus with uu, vv, and ww.