Half-planes in $\mathbb{R}^2$

Question

I proved a proposition in $\mathbb{R}^2$ using a cartersian coordinate system. Now I want to provide another proof of the same proposition without the cartesian coordinate system. Anyway I need to express some half-planes through the scalar product. My skill in geometry are very basic and I'm having some difficulties. I try to explain the matter: if $x_0,x_1\in\mathbb{R}^2$ (with $x_0\neq x_1$), the set $\Delta=\{x\in\mathbb{R}^2\ \ :\ \ \langle x-x_0,x_1-x_0\rangle\ge 0\}$ is the half-plane containing $x_1$ whose boundary is orthogonal to the line $r$ passing through $x_0$ and $x_1$. These are the questions:

1. Fixed an angle $\alpha$, there is a way to write the vectorial equations of the two lines $r_1$ and $r_2$, passing through $x_0$, inclined at an angle $\alpha$ with respect to the line $r=\{x_0+\lambda (x_1-x_0)\ \ :\ \ \lambda\in\mathbb{R}\}$?
2. There is a way to define the two half-planes (having $r_1$ and $r_2$ as boundary) using a similar expression of the set $\Delta$, eventually using also the norms of the involved vectors?

Answer 1

First, you have to answer the question : what is the mathematical definition of an half-plane? Or even what is a half-space? Generalizing, it won't be much harder to answer.

Let $E$ be a real vector space (for example, $\mathbb R^2, \mathbb R^3, \mathbb R^4,...)$ . Then let $f\neq0$ be a linear form on $E$. And let $a\in \mathbb R$. $$H:=\{x\in E: f(x)=a\}$$is an hyperplane of E. In $\mathbb R^2$, the hyperplanes are the lines.

For example let $E:=\mathbb R^2, x_0\in E$ with its Euclidean structure, where the dot product is denoted $\langle.,.\rangle$, $x_1\in E$, $a=\langle x_0,x_1-x_0\rangle$ and $f:E\to \mathbb R, x\mapsto \langle x,x_1-x_0\rangle$. $H$ is the line passing through $x_0$ and perpendicular to the line $r$ that you defined. Now let's come to the mathematical definition of a half-space : $$\{x\in E: f(x)\geq a\}\text{ and }\color{orange}{\{x\in E: f(x)\leq a\}}$$are called closed half-spaces containing H.

In a second step, you can use your point 1 to convince yourself that this definition corresponds to your intuition by using $$\cos \alpha=\frac{\langle u,v\rangle}{||u||.||v||}$$where $\alpha:=\widehat{(u,v)} $ is such that $-\frac{\pi}{2} \leq\alpha \leq \frac{\pi}{2}$