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 Rn.
be the set of points equidistant from u and v,
and let m≜2u+v 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.
Proof. Let m+w∈m+span(v−u)⊥, so that ⟨w,v−u⟩=0.
Note that w and m−u are orthogonal, as
Similarly, w and m−v are orthogonal. As m−u=−(m−v),
we have ∥m−u∥=∥m−v∥. Then, with
the Pythagorean theorem,
so m+w is equidistant from u and v, and thus,
Decomposition of the distance from u [or v] to our given
vector, m+w, into orthogonal parts.
Conversely, suppose w∈A, so that ∥w−u∥=∥v−u∥.
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
Therefore, we have
so u+v−2w is orthogonal to v−u. Furthermore,