Matrix methods for equation of a line

[zcd]

Given two points on R² how would one find the constants a,b,c such that
ax+by+c=0 gives the line crossing the two points (with matrix methods)?

 

[Simon_Tyler]

So you're given the points $\vec{p}_1=(x_1, y_1)$ and $\vec{p}_2=(x_2, y_2)$.

A line is given by the equation

$a x + b y =\begin{pmatrix} a & b \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} = c$

Which, assuming $c \neq 0$, can be rescaled to
$\begin{pmatrix} a & b \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} = 1$
Then the two points must satisfy
$\begin{align} &\begin{pmatrix} a & b \end{pmatrix}\cdot\begin{pmatrix}x_1&x_2 \\ y_1&y_2\end{pmatrix} = \begin{pmatrix} 1 & 1 \end{pmatrix} \\ \implies &\begin{pmatrix} a & b \end{pmatrix} = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix}x_1&x_2 \\ y_1&y_2\end{pmatrix}^{-1} =\frac{1}{x_1 y_2 - x_2 y_1}\begin{pmatrix} y_1-y_2 & x_1-x_2\end{pmatrix} \end{align}$
And so we have the equation for the line. (This is just "[URL rule[/URL])

Note that if $\det(\vec{p}_1, \vec{p}_2) = x_1 y_2 - x_2 y_1 = 0 \,,$ (which happens when $\vec{p}_1\propto\vec{p}_2$)
then the above does not make sense and the line must go through the origin, i.e. $c=0$.
In which case,
$a x = - b y \quad \implies \quad y = -\frac{a}{b}x$
and we can just use either point to find the single parameter determining the line
$\frac{a}{b} = -\frac{y_1}{x_1} = -\frac{y_2}{x_2} \ .$