Why $ax+by=c$ divides the plane in two half planes characterized by inequalities

Question

The line $ax+by=c$ divides the plane into two half planes given by $ax+by<c$ and $ax+by>c$. My question is why all points in the any fixed half plane will satisfy exactly one of these inequalities.

Answer 1

The function $f(x,y)=ax+by-c$ is positive on one side of the line and negative on the other by the intermediate value theorem.

If $f(x_1,y_1)$ is negative and $f(x_2,y_2)$ is positive, then the value of $f$ on the line segment joining $A=(x_1,y_1)$ and $B=(x_2,y_2)$ must be zero somewhere, so the line segment must cross $ax+by=c$.

(Given that everything is "linear," you can actually find the point on the line between A and B that hits $ax+by=c$, so you don't technically need IVT, but IVT shows the same thing about more general functions $f$ - if you subtract the points where $f(X)=0$, then the remaining points will partition into "connected components" on which $f$ is either strictly positive or strictly negative.)

Answer 2

Can you understand why every point on a one dimensional numberline is either greater than or less than (or equal to) a defined point on that numberline?

If so, then imagine the $xy$ plane as just an infinite collection of vertical numberlines. And the dividing point on each numberline is defined by some function of $x$. So every point on the plane falls on one (and only one) of these numberlines and must be on one side or the other of it.