Intuition about normalvector and equation if a line in the plane

Question

The general form for an equation of a line in the $(x,y)$ plane is

$$ a*x+b*y = c $$

Then, the vector $(a,b)$ is perpendicular to it.

I accept it as a fact, but how may I develop an intuition about this?

Answer 1

Let $M(x,y)$ be a point in the plane, $\vec n$ the vector with coordinates $(a,b) $. If $M_0(x_0, y_0)$ is any other point in the plane, the equation of the plane can be written as $$ax+by=ax_0+by_0\iff a(x-x_0)+b((y-y_0)=0\iff \vec n\cdot \overrightarrow{M_0M}=0.$$

Answer 2

Consider the vector $\mathbf v=(a,b)$ and a line parallel to it and passing through the origin.
On that line, starting from the origin mark the points $0\mathbf v,\;1\mathbf v,\;2\mathbf v,\; \cdots$ and negatives.
A second line, normal to the first and thus to $\mathbf v$, and crossing it at $\lambda \mathbf v$, is the locus of all the points $\mathbf p$ for which $\mathbf p \cdot \mathbf v=\lambda\; \mathbf v \cdot \mathbf v$.

Answer 3

Recall that the slope of a line $g$ through the origin and $(a,b)$ is $b/a$, provided that $a\neq0$. Now the slope of the line $h$ given by $ax+by=c$ is $-a/b$, provided that $b\neq0$. As the product of the slope is $-1$, we conclude that $g\perp h$. (The intuition of the missing cases is obvious.)