How can we derive the standard form of the linear equation: $Ax+By+C$?

Question

How can we derive the standard form of the linear equation: $Ax+By+C=0$? What do "$A$", "$B$" and "$C$" in the standard form of the linear equation mean? As in the point-slope form of the linear equation: $y-y_0=m(x-x_0)$, $m$ is the slope, and $x$, $y$ are the coordinates of any point. Likewise, $x_0$, $y_0$ are the coordinates of the given point.

Answer 1

Let's say we start with the point-slope form $$y-y_0=m(x-x_0),$$ where $(x, y)$ represents any point on the line, and $(x_0, y_0)$ represents a given point on the line, and $m$ is the given slope of the line. Now, $$ \begin{align} y-y_0 =m(x-x_0) & \iff y - y_0 = mx - mx_0 \\ \\ & \iff y - y_0 - mx - mx_0 = 0 \\ \\ & \iff (-m)x + y + (-y_0 - mx_0) = 0 \end{align} $$

Now, we can simply let $A = -m,\; B = 1,\; C = (- mx_0 - y_0)$, which means that $$y + (-m)x + (-y_0 - mx_0) = 0 \iff Ax + By + C = 0$$

Answer 2

If you interpret it as a vector equation, it makes a lot of sense...

$$ \begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{V}\cdot\mathbf{r} = -C $$ where $\mathbf{r}$ is the position vector and $\mathbf{V}$ is a vector containing the two coefficients. This can be interpreted as "the set of points $\mathbf{r}$ whose length in the direction of $\mathbf{V}$ is $\frac{-C}{|\mathbf{V}|}$". That is, $\mathbf{V}$ is a vector perpendicular to the line. If you divide through by $\sqrt{A^2+B^2}$, you find that $|\mathbf{V}|=1$ and $\mathbf{V}$ is the unit vector normal to the line, while $-C$ is the distance from the origin to the point on the line closest to the origin.

This works in 3D, as well, where you now have $\mathbf{V}$ being the unit vector normal to the plane.