Show that the set of points $\left\{x\in\mathbb{R}^2 \mid \frac{(x-p)\cdot(q-p)}{|q-p|}\ < \frac{1}{2}|q-p|\right\}$ is the half plane.

Question

Let $H(p,q)$ be a half plane in $\mathbb{R}^2$ such that $H(p,q) = \{x\in\mathbb{R}^2 \mid |x - p| < |x-q|\}$.

Show that

$$H(p,q) = \left\{x\in\mathbb{R}^2 \mid \frac{(x-p)\cdot(q-p)}{|q-p|}\ < \frac{1}{2}|q-p|\right\}$$

My attempt so far

$$|x-p| < |x-q|$$

$$\sqrt{|x|^2+|p|^2-2x\cdot p} < \sqrt{|x|^2+|q|^2-2x\cdot q}$$

$$|x|^2+|p|^2-2x\cdot p < |x|^2+|q|^2-2x\cdot q$$

$$-2x\cdot p+2x\cdot q < |q|^2-|p|^2$$

$$-2x\cdot p+2x\cdot q < q\cdot q-p\cdot p$$

$$-2x\cdot(q-p) <(q+p)\cdot(q-p)$$

But i feel like I've reached a dead end. Probably because I took the wrong direction/approach. Any tips to solving the stated problem?

Answer 1

Pick up at this line: $$|x|^2+|p|^2-2x\cdot p < |x|^2+|q|^2-2x\cdot q$$

$$2x \cdot q - 2x \cdot p + |p|^2 < |q|^2$$

$$2x \cdot q - 2x \cdot p - 2p \cdot q + 2|p|^2 < |q|^2 - 2 p \cdot q + |p|^2$$

$$2(x - p) \cdot (q-p) < (q-p) \cdot (q-p)$$

I think it is easy enough to finish from here.

Answer 2

$$\begin{align*}&|x-p|^2<|x-q|^2\iff (x-p)\cdot (x-p)<(x-q)\cdot(x-q)\iff\\{}\\ &|x|^2-2x\cdot p+|p|^2<|x|^2-2x\cdot q+|q|^2\iff|p|^2+2x\cdot(q-p)<|q|^2\iff\\{}\\ &|p|^2+2(x-p)\cdot(q-p)<|q|^2-2p\cdot(q-p)<0\iff\\{}\\ &2(x-p)\cdot(q-p)<|q|^2-2p\cdot q+|p|^2=|q-p|^2\end{align*}$$