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I need to solve an equation of a line using three known coordinate pairs (x0, y0), (x1, y1), and (x2, y2).

The equation of the plane is, of course, ax + by + c = 0.

I'm writing a little piece of code to calculate the position of a point w.r.t a line as it changes, so I just want to reduce the math as much as possible for efficiency, which means pre-solving this in terms of the x and y coordinates.

I can derive the equations to solve for a, b and c by hand in advance and write the code that way, but it ends up an ugly mess of substitutions to do it.

Is there a more straightforward approach that, giving my coordinates, I can solve for the constants quickly, without having to rely on matrix math or pulling in a math library?

Joel Graff
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  • In what language are you coding? Is it a high level one like Python? Many of such scripting languages have built in linear solvers that you can use. (just saying) – Cardinal Jul 06 '20 at 20:43
  • BTW, the easiest way is forming the matrix representation of the equations. – Cardinal Jul 06 '20 at 20:46
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    $ax + b y + c$ is a line, not a plane. If your points are in three-dimensional space, they need three coordinates, not two. – Robert Israel Jul 06 '20 at 20:48
  • @Cardinal: using a linear solver for such tiny problems is complete overkill. –  Jul 06 '20 at 20:49
  • @YvesDaoust I know it was just a suggestion, I would go with a nice inverse matrix! 0:-) – Cardinal Jul 06 '20 at 20:50
  • The plane (in three-dimensional space) through three non-collinear points $A,B,C$ is orthogonal to the vector cross product $(A-B) \times (C-B)$. – Robert Israel Jul 06 '20 at 20:50
  • Mh, a line by three points ? –  Jul 06 '20 at 20:52
  • @YvesDaoust Hmm, depends on the algorithm though, but I see what you say :) You are right. If you want to be lazy then ... :embarrassed: – Cardinal Jul 06 '20 at 20:52
  • @Robert Israel: my bad on the equation of a plane reference and the lines are coplanar, so the z-coordinate is negelected. Anyway, I'm sorting out two problems at once at the moment, so no surprise I confused my terms. lol – Joel Graff Jul 06 '20 at 20:53
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    @Cardinal: I am using Python, but it doesn't come with linear solvers - I'd have to add scipy or numpy packages, which I'm working very hard to avoid - to the extent I'm re-implementing vector math in tuple form just so I don't have to depend on numpy. – Joel Graff Jul 06 '20 at 20:54
  • @YvesDaoust The laziness algorithm as I said :-) – Cardinal Jul 06 '20 at 20:55

1 Answers1

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A line by two points is

$$\begin{vmatrix}x&y&1\\x_0&y_0&1\\x_1&y_1&1\end{vmatrix}= \begin{vmatrix}y_0&1\\y_1&1\end{vmatrix}x- \begin{vmatrix}x_0&1\\x_1&1\end{vmatrix}y+ \begin{vmatrix}x_0&y_0\\x_1&y_1\end{vmatrix}=0.$$

A plane by three points is

$$\begin{vmatrix}x&y&z&1\\x_0&y_0&z_0&1\\x_1&y_1&z_1&1\\x_2&y_2&z_2&1\end{vmatrix} =\begin{vmatrix}y_0&z_0&1\\y_1&z_1&1\\y_2&z_2&1\end{vmatrix}x-\begin{vmatrix}x_0&z_0&1\\x_1&z_1&1\\x_2&z_2&1\end{vmatrix}y+\begin{vmatrix}x_0&y_0&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}z-\begin{vmatrix}x_0&y_0&z_0\\x_1&y_1&z_1\\x_2&y_2&z_2\end{vmatrix} =\begin{vmatrix}y_{10}&z_{10}\\y_2&z_{20}\end{vmatrix}x-\begin{vmatrix}x_{10}&z_{10}\\x_{20}&z_{20}\end{vmatrix}y+\begin{vmatrix}x_{10}&y_{10}\\x_{20}&y_{20}\end{vmatrix}z-\begin{vmatrix}x_0&y_0&z_0\\x_{10}&y_{10}&z_{10}\\x_{20}&y_{20}&z_{20}\end{vmatrix} =0$$

where $a_{ij}:=a_j-a_i$.