I was trying to prove the formula for distance of a point in the cartesian plane from a line. And there are many easy proofs. I was looking for something “tastier”. For equations of planes in 3d, the coefficients $A$, $B$ and $C$ in $Ax + By + Cz + D = 0$ represent a vector perpendicular to the plane! My goal was to find such a connection for 2d lines. Indeed, for $Ax + By = 0$, the dot product of any vector $(x,y)$ with $(A,B)$ is $0$. And the constant term can arise when we shift our origin!
My goal then is to see how this shift of the origin relates to the constant term that’s added. Notice that the intercepts of the line $Ax + By + C = 0$ are $|\frac{-C}{A}|$ and $|\frac{-C}{B}|$ and the distance between this line and another line $Ax + By + D = 0$ is $\frac{\left|C-D\right|}{\sqrt{A^{2\ }+B^{2}}}$. The pattern is clearly visible if we set $D = 0$. I wanted to see how I could relate this to vectors. Firstly, the distance between the origin and the line can be written as $k\sqrt{A^2 + B^2}$ where $k \in \Bbb{R^+}$. This distance is also the dot product $(kA,kB)\cdot(|\frac{-C}{A}|,0)$. Hence, $|kC|$ is the same distance. Notice that the line equation $Ax + By + C = 0$ can be written as $A(x-kA) + B(y-kB) = 0$. It is clear that $-C = kA^2 + kB^2$. Evaluating all previous equations, we get $$\frac{|C|}{A^2 + B^2}\sqrt{A^2 + B^2} = |kC|$$
I want to expand this further, can this be done along more vectors other than $(1,0)$ or $(0,1)$ or $(A,B)$? My understanding is of a highschool student but dont dumb your answers down, but don’t assume my knowledge of linear algebra or algebraic geometry to be like a math major’s. (I will probably join a BMath course soon though!)
Let $(x_0,y_0)$ be any point on the line in the Cartesian plane. Then given $(A,B)$ represents a normal vector to the line, we may write an equation for the line as
$$(A,B)\cdot (x-x_0,y-y_0)=0\iff Ax+By=Ax_0+By_0=(A,B)\cdot(x_0,y_0).$$
So when we write the equation of a line as $$Ax+By=C,$$
the constant $C$ represents the dot product between the chosen normal vector $(A,B)$ and any point on the line.
It is easiest to interpret the constant if we choose $(A,B)$ as a unit normal vector pointing away from the origin, so that by the dot product formula,
$$(A,B)\cdot (x_0,y_0)=\|(A,B)\|\|(x_0,y_0)\|\cos \theta=\|(x_0,y_0)\|\cos \theta\quad (1)$$
where $\theta$ is the angle between $(A,B)$ and $(x_0,y_0)$. Note $0\leq \theta\leq 90^\circ$ since we chose the normal vector pointing away from the origin. Now you can see quite clearly that the constant on the right-hand side of $(1)$ simply represents the length of the vector projection of $(x_0,y_0)$ on $(A,B)$, denoted by $\text{proj}_{(A,B)}(x_0,y_0)$, illustrated below. This length is exactly the distance between the origin and the line.