Computing the Intersection of Two Lines

Two lines $ax + by + c = 0$ and $dx + ey + f = 0$ intersect when $ax + by + c = dx + ey + f$ . We can solve for $x$ and $y$ in terms of the coefficients as follows. First, solve for $x$ :

(1) $\begin{align*}ax + by + c &= 0\\ax &= -by - c\\ x &= \frac{-by - c}{a}.\end{align*}$

Second, plug the solution into the second line equation:

(2) $\begin{align*}d\left(\frac{-by - c}{a}\right) + ey + f &= 0\\ \frac{-bdy - cd}{a} + ey + f &= 0\\ -bdy-cd+aey+af &= 0\\ aey-bdy &= cd-af\\ y(ae-bd) &= cd-af\\ y &= \frac{cd-af}{ae-bd}.\end{align*}$

Repeat this process, solving for $y$ first, to obtain an expression for $x$ :

(3) $\begin{align*}ax + by + c &= 0\\by &= -ax - c\\ y &= \frac{-ax - c}{b}.\end{align*}$

Plug the solution into the second line equation:

(4) $\begin{align*}dx + e\left(\frac{-ax - c}{b}\right) + f &= 0\\ dx + \frac{-aex - ce}{b} + f &= 0\\ bdx-aex-ce+bf &= 0\\ bdx-aex &= ce-bf\\ x(bd-ae) &= ce-bf\\ x &= \frac{ce-bf}{bd-ae}.\end{align*}$

Note that the denominators of both coordinates are the same modulo sign, so we can multiply either by $-1/-1$ to normalize the denominator and make its calculation reusable (below we apply this to $y$ ). This leaves us with the final expression:

(5) $\begin{align*}(x, y) &= \left(\frac{ce-bf}{bd-ae}, \frac{af-cd}{bd-ae}\right).\end{align*}$

The following demo allows you to specify two lines via two points each and observe the intermediate intersection point calculations.

intX numerator:
intX denominator:
intY numerator:
intY denominator:

Clinton Freeman

Computer Science, Design

Computing the Intersection of Two Lines

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