Equation of Line with Determinant¶

Let's say that we want the equation of a line going through point $P$ in the direction of a nonzero vector $\vec d$. We know that the slope of the line is $\frac{d_y}{d_x}$, so by plugging into this result, we get $$ y - P_y = \frac{d_y}{d_x}(x - P_x). $$ This method doesn't work if $d_x = 0$; that is, when $\vec d$ is vertical. In that case, the line is also vertical, so its equation is simply $x = P_x$. On this page, we'll use determinants to find the line equation in a way that works the same in both cases.

We see that a point $(x,y)$ is on the line if and only if the vector from $P$ to $(x,y)$ is in the direction of the line.

Let $\vec d$ be any nonzero vector going in the direction of the line. Now the vector from $P$ to $(x,y)$ goes in the direction of the line if and only if it's collinear with $\vec d$. That's true if and only if the parallelogram created by $\vec d$ and the vector from $P$ to $(x,y)$ is "flat".

The parallelogram is "flat" if and only if the determinant with the vectors as rows is zero. The vector from $P$ to $(x,y)$ is $(x-P_x)\I + (y-P_y)\J$. From all this, we see that $(x,y)$ is on the line if and only if $$ \det\begin{bmatrix} x-P_x & y-P_y \\ d_x & d_y \end{bmatrix} = 0. $$

If we know two points of the line, $P$ and $Q$, then we can choose $$ \vec d = \bigvec{PQ} = (Q_x-P_x)\I + (Q_y-P_y)\J. $$ This must not be the zero vector, so we need $P \ne Q$.

The line going through $P$ and $Q$ is same as the line going through $Q$ and $P$. This means that in the above result, it doesn't matter which point is $P$ and which is $Q$.