Half Spaces

The equation $3x-4y=24$ is the so-called “standard form” equation of a line.

The inequality $3x-4y<24$ represents a region of the plane called a “half-space”, and the points that satisfy the inequality constitute the entire half of the plane which lies to one side of the line.

A geometric approach

Let $N=\left<3,-4\right>$ be a vector. We’ll show its direction is perpendicular to the line $3x-4y=c$.

Suppose $p_0=(x_0,y_0)$ is any fixed point on the line, so $3x_0 - 4y_0 = c$. Then let $p=(x,y)$ be some other point. Now $p$ is on the line if and only if $N \cdot p = c$. Equivalently, $N \cdot (p-p_0) = N \cdot p - N \cdot p_0 = 0$. But the equation $N \cdot (p-p_0)$ holds exactly when the displacement $p-p_0$ is a vector perpendicular to $N$.

So the points $p$ which satisfy $N \cdot p = c$ form a line perpendicular to $N$.

Since the ray $R_N = \left\{ \lambda N | \lambda \in \mathbb{R} \right\}$ is perpendicular to the line, we can find a point $p_0 \in R_N$ which lies on the line. Solving for $\lambda$ we have: $N \cdot (\lambda N) = c \longrightarrow \lambda = c/\Vert N \Vert^2$ and $p_0 = \lambda N$ satisfies $N \cdot p_0 = c$. It is also the closest point on the line to the origin and has magnitude $\frac{\vert c\vert}{\Vert N\Vert}$.

So changing $c$ in the equation $N \cdot p = c$ moves the solution set (line) forwards or backwards along $N$ to a new line whose distance to $0$ is proportional to $\vert c \vert$.

Inequality

Define the function $f(p) = N \cdot p = 3x-4y$. We know $p_0$ is a solution to $f(p) = c$ and the other solutions form a line perpendicular to $N$ which passes through $p_0$. Since $f$ is computed as a dot product, the value of $f(p)$ is proportional to the length of the projection of $p$ along the $N$ direction. Also, $f(p)$ is negative when the projection points in the opposite direction of $N$. In particular $f(-\lambda N)<0$ for every $\lambda>0$.

To know which points satsify $f(p) < c$ simply look at all the points which satisfy $f(p) = r$ for $r \in (-\infty, c)$

The solutions of $f(p)=r$ form a line $L_r$ perpendicular to $N$ passing through $p_0(r)=r N / \Vert N\Vert^2$ with equation $N \cdot p = r$ so the contours of $f$ are the lines $\{L_r : r \in \mathbb{R}\}$.

The half-space $f(p)<c$ is generated by the lines $\{L_r | -\infty < r < c \}$. Equivalently, if $p_0$ is any point on the line $f(p) = c$ then the half-space $f(p) < c$ consists of all points $p$ whose projections in the $N$ direction are “not as far” along $N$ as $p_0$ achieves. So starting anywhere on the line and moving along $-N$ gives a point in the half-space $f(p)<c$.