Why is this set a half plane?

Question

The set $$\{ x: \| x - x_0\| \leq \|x-a\| , x\in \mathbb{R}^2\}$$

Where we fix the element $a$ and $x_0$ in $\mathbb{R}^2$.

I don't see how this is equivalent to

$$\{ x: A^t x \leq b\}$$ for some $A\in \mathbb{R}^2$ and $b\in \mathbb{R}$.

Answer 1

If you square both sides (nothing is lost as both sides are already non-negative) and cancel things, you get $$ \|x\|^2+\|x_0\|^2-2x_0^tx\leq\|x\|^2+\|a\|^2-2a^tx, $$ so $$ a^tx-x_0^tx\leq\frac12\,\left(\|a\|^2-\|x_0\|^2\right), $$ or $$ (a-x_0)^tx\leq\frac12\,\left(\|a\|^2-\|x_0\|^2\right). $$

Answer 2

You can see it geometrically as follows. If $a,b$ are distinct points in the plane, it's pretty clear that the set of points that are equidistant from $a$ and $b$ is the line that bisects the segment joining $a$ and $b$. This means the points on one side are closer to $a$ and the points on the other side are closer to $b$.