## Introduction

In two-dimensions, given two lines each defined by two points you can usually find a point where the lines intersect. Wikipedia gives the coordinates1 in terms of determinants, but not a derivation.

## Derivation Suppose $$\textbf{a}$$ and $$\textbf{b}$$ are on the first line, and $$\textbf{c}$$ and $$\textbf{d}$$ are on the second. Then, assuming that the lines aren’t parallel, we can write a general point as

$\textbf{x} = \mu \, (\textbf{b} - \textbf{a}) + \nu \, (\textbf{d} - \textbf{c}).$

So to find $$\textbf{x}$$ we just need to find $$\mu$$ and $$\nu$$. We can get rid of $$\nu$$ by taking the dot-product with $$\textbf{d} - \textbf{c}$$ rotated by $$\pi / 2$$, but the notation gets a bit messy. Equivalently we can embed the 2D vectors in the $$z = 0$$ plane of a 3D space, then look at the cross-product with $$\textbf{d} - \textbf{c}$$:

$(\textbf{d} - \textbf{c}) \times \textbf{x} = \mu \, (\textbf{d} - \textbf{c}) \times (\textbf{b} - \textbf{a}).$

Now, $$\textbf{x}$$ lies between $$\textbf{c}$$ and $$\textbf{d}$$, so

\begin{align} \left(\textbf{d} - \textbf{c}\right) \times \left(\textbf{x} - \textbf{d}\right) &= 0,\\ \left(\textbf{d} - \textbf{c}\right) \times \textbf{x} - \left(\textbf{d} - \textbf{c}\right) \times \textbf{d} &= 0,\\ \left(\textbf{d} - \textbf{c}\right) \times \textbf{x} &= \left(\textbf{d} - \textbf{c}\right) \times \textbf{d},\\ \left(\textbf{d} - \textbf{c}\right) \times \textbf{x} &= \textbf{d} \times \textbf{c}. \end{align}

Which we can substitute to give

$\mu \, (\textbf{d} - \textbf{c}) \times (\textbf{b} - \textbf{a}) = \textbf{d} \times \textbf{c}.$

The results of both cross-products lie along $$\textbf{z}$$, so we divide them with the understanding that we’re just looking at the $$z$$-components,

$\mu = \frac{\textbf{d} \times \textbf{c}}{(\textbf{d} - \textbf{c}) \times (\textbf{b} - \textbf{a})}.$

By a similar analysis we also have,

$\nu = \frac{\textbf{b} \times \textbf{a}}{(\textbf{b} - \textbf{a}) \times (\textbf{d} - \textbf{c})}.$

And thus,

$\textbf{x} = \frac{\textbf{d} \times \textbf{c}}{(\textbf{d} - \textbf{c}) \times (\textbf{b} - \textbf{a})} \, (\textbf{b} - \textbf{a})- \frac{\textbf{b} \times \textbf{a}}{(\textbf{d} - \textbf{c}) \times (\textbf{b} - \textbf{a})} \, (\textbf{d} - \textbf{c}).$

## Assumptions

The main assumption is that the cross-product in the denominator doesn’t vanish, i.e.,

$(\textbf{b} - \textbf{a}) \times (\textbf{d} - \textbf{c}) \neq 0,$

which is equivalent to saying that the lines aren’t parallel. With finite precision arithmetic though, one needs also to beware the case where the lines are almost parallel.

## Implementation

Here’s a trivial implementation in stand-alone Haskell:

-- Our vector
type V2 = (Double,Double)

-- Subtraction
(.-.) :: V2 -> V2 -> V2
(xa,ya) .-. (xb,yb) = (xa - xb, ya - yb)

-- Scalar multiplication
(*.) :: Double -> V2 -> V2
a *. (x,y) = (a * x, a * y)

-- z-component of cross-product
(.^.) :: V2 -> V2 -> Double
(xa,ya) .^. (xb,yb) = xa * yb - xb * ya

intersect :: V2 -> V2 -> V2 -> V2 -> V2
intersect a b c d = (s *. ba) .-. (t *. dc)
where ba = b .-. a
dc = d .-. c
z  = dc .^. ba
s  = d .^. c / z
t  = b .^. a / z