Equation of a line passing through 2 points

There are many ways to write this equation. Using the traditional slope-intercept equation, $y = mx + y_0$ , we have $m = \frac{\Delta y}{\Delta x} = \frac{d - b}{c - a};\ \ y_0 = b - ma = \frac{b(c-a) - (d - b)a}{c - a} = \frac{bc - da}{c - a}, \\ y = \frac{(d-b)x + bc - da}{c - a}.$ Or, use the fact that any two points on the line should give the same slope value: $\frac{y - b}{x - a} = \frac{d - b}{c - a};\ \ \ y = b + \frac{(x - a)(d - b)}{c-a},$ which is easily simplified to the same equation.

Or write $(x,y)$ as an affine combination, $(x,y) = \lambda(a,b) + \mu(c,d)$ with $\lambda + \mu = 1$ ; then $\left[\begin{array}{ccc} a & c & x \\ b & d & y \\ 1 & 1 & 1 \end{array}\right]\left(\begin{array}{c}\lambda \\ \mu \\ -1\end{array}\right) = \left(\begin{array}{c}0 \\ 0 \\ 0\end{array}\right)$ must have a non trivial solution, so that $0 = \left|\begin{array}{ccc} a & c & x \\ b & d & y \\ 1 & 1 & 1 \end{array}\right| = x(b - d) + y(c - a) + (ad - bc),$ so that $y = -\frac{x(b - d) + (ad - bc)}{c - a} = \frac{x(d-b) + (bc - ad)}{c-a}.$