# 2D Line Mathematics Using Homogeneous Coordinates

## Introduction

It is possible to compute the intersection of two lines and the line given two points using cross-product.

In this blog post, I would like to quickly derive how to do so using homogeneous coordinate representations.

## 2D Point Representations

### Inhomogeneous Coordinates

The inhomogeneous coordinates for a 2D point are just ordinary two-value Cartesian coordinates.

$$\mathbf{x} = (x, y)$$

### Augmented Coordinates

The augmented coordinates for a 2D point are just the 2D inhomogeneous coordinates with an additional constant $1$.

$$\bar{\mathbf{x}} = (x, y, 1)$$

### Homogeneous Coordinates

The homogeneous coordinates are just the augmented coordinates scaled by some value $\tilde{w}$.

\begin{align} \tilde{\mathbf{x}} &= \tilde{w} \bar{\mathbf{x}} \\ &= \tilde{w} (x, y, 1) \\ &= (\tilde{w}x, \tilde{w}y, \tilde{w}) \\ &= (\tilde{x}, \tilde{y}, \tilde{w}) \\ \end{align}

where $\tilde{w} \in \mathbb{R}$.

When $\tilde{w} = 0$, $\tilde{\mathbf{x}}$ is called ideal point and do not have the corresponding inhomogeneous coordinates.

## Intersection

The 2D line $l: ax + by + c = 0$ could be represented using homogeneous coordinates, $\tilde{\mathbf{l}} = (a, b, c)$. It can also be normalized so that $\mathbf{l} = (\hat{n}_x, \hat{n}_y, d) = (\mathbf{n}, d)$ with $\left\Vert \mathbf{n} \right\Vert = 1$.

Suppose $\mathbf{x} = (x, y)$ is the intersection of two lines $\tilde{\mathbf{l}}_1 = (a_1, b_1, c_1)$ and $\tilde{\mathbf{l}}_2 = (a_2, b_2, c_2)$, we must have

$$\begin{gather} \bar{\mathbf{x}} \cdot \tilde{\mathbf{l}}_1 = 0 \\ \bar{\mathbf{x}} \cdot \tilde{\mathbf{l}}_2 = 0 \\ \end{gather}$$

Because the cross product of $\tilde{\mathbf{l}}_1$ and $\tilde{\mathbf{l}}_2$, $\tilde{\mathbf{l}}_1 \times \tilde{\mathbf{l}}_2$, is perpendicular to both $\tilde{\mathbf{l}}_1$ and $\tilde{\mathbf{l}}_2$, i.e.,

$$\begin{gather} (\tilde{\mathbf{l}}_1 \times \tilde{\mathbf{l}}_2) \cdot \tilde{\mathbf{l}}_1 = 0 \\ (\tilde{\mathbf{l}}_1 \times \tilde{\mathbf{l}}_2) \cdot \tilde{\mathbf{l}}_2 = 0 \\ \end{gather}$$

We must have

\begin{align} \tilde{\mathbf{l}}_1 \times \tilde{\mathbf{l}}_2 &= \tilde{w} \bar{\mathbf{x}} \\ &= (\tilde{w}x, \tilde{w}y, \tilde{w}) \\ \end{align}

for some $\tilde{w} \in \mathbb{R}$.

Therefore,

\begin{align} \bar{\mathbf{x}} = &= \frac{1}{\tilde{w}} \tilde{\mathbf{l}}_1 \times \tilde{\mathbf{l}}_2 \\ &= \frac{1}{\tilde{w}} (\tilde{w}x, \tilde{w}y, \tilde{w}) \\ \end{align}

## 2D Line from 2D Points

Suppose the line $\tilde{\mathbf{l}} = (a, b, c)$ passes two points $\mathbf{x}_1 = (x_1, y_1)$ and $\mathbf{x}_2 = (x_2, y_2)$, similar to the intersection calculation, we must have

$$\begin{gather} \bar{\mathbf{x}}_1 \cdot \tilde{\mathbf{l}} = 0 \\ \bar{\mathbf{x}}_2 \cdot \tilde{\mathbf{l}} = 0 \\ \end{gather}$$

Because the cross product of $\bar{\mathbf{x}}_1$ and $\bar{\mathbf{x}}_2$, $\bar{\mathbf{x}}_1 \times \bar{\mathbf{x}}_2$, is perpendicular to both $\bar{\mathbf{x}}_1$ and $\bar{\mathbf{x}}_2$, i.e.,

$$\begin{gather} \bar{\mathbf{x}}_1 \cdot (\bar{\mathbf{x}}_1 \times \bar{\mathbf{x}}_2) = 0 \\ \bar{\mathbf{x}}_2 \cdot (\bar{\mathbf{x}}_1 \times \bar{\mathbf{x}}_2) = 0 \\ \end{gather}$$

We must have

\begin{align} \bar{\mathbf{x}}_1 \times \bar{\mathbf{x}}_2 &= \tilde{v} \bar{\mathbf{l}} \\ &= \tilde{\mathbf{l}} \\ \end{align}

If the two points were represented using homogeneous coordinates, equivalently,

\begin{align} \tilde{\mathbf{x}}_1 \times \tilde{\mathbf{x}}_2 &= \tilde{\mathbf{l}} \\ \end{align}

Lei Mao

11-01-2021

11-01-2021