the set of points equidistant from $ u $ and $v$ form a line.

Question

Let $u$ and $v$ be two vectors in $ \mathbb{R}^2 $ with the standard norm. Show that the set of points equidistant from $ u $ and $v$ form a line.

I show that if $x$ is equidistant from $u$ and $v$, then for all $\gamma$ $\in$ $ \mathbb{R} $ $\gamma $$x$ to be equidistant from $ u $ and $v$.$\gamma $$x$ form a line. Unfortunately I can't show any vector $a$ that is equidistant from $ u $ and $v$ in the line $\gamma$$x$.

Answer 1

Note that you don't want equal constant distance, but equal normal distance at any point. At this point of time it is more beneficial to consider how equations of angle bisector(s) are derived in analytic geometry.

but how can i say all points that is equidistant from u and v will be on this line ? –rmznyzgyr

EDIT1: @rmznyzgyr: Good question. Remember, distance between a point and a line refers only to the minimum distance between them, but not any or all distances. This terminology is a geometrical convention.

EDIT2:

Equal Normal Distance

DE = DF at any point on the internal (or external) bisector of given angle included between u and v.

For unit vectors $u = e^\alpha $ and $v = e^\beta$ ... the vector $ e^{(\alpha+\beta)/2} $ is the bisector.

Answer 2

Difference of squares! \begin{align*} \|x-u\|=\|x-v\| &\iff \|x-u\|^2 - \|x-v\|^2 = 0 \\ &\iff \langle x-u+x-v, x-u-x+v\rangle = 0 \\ &\iff \langle x-\tfrac{u+v}{2}, v-u\rangle = 0 \end{align*} So $x$ is equidistant from $u$ and $v$ if and only if the vector from $\frac{u+v}{2}$ to $x$ is perpendicular to the vector from $u$ to $v$. At this point you can supply a geometric argument that the set of such $x$ forms a line; alternatively, you can rewrite the above condition as $$ (v-u)^T x = (v-u)^T\tfrac{u+v}{2} $$ which is a matrix equation $Ax=b$ with $A$ being $1\times 2$ and not zero (assuming $u\ne v$). General considerations about linear systems will tell you that the solutions of such an equation form a line.