Prove that all points on the same side of a straight line yields only a positive or negative value.

Question

Can I prove that $f(x,y)=ax+by+c$ yields only positive or only negative values, for any two points on the same side of the straight line $ax+by+c=0$? If so, how?

Answer 1

On the line, $x = \frac{by+c}{a}$. Take the side where $x <\frac{by+c}{a}$. Then $$f(x,y) = ax+by+c > 2ax$$ and $$f(x,y) < 2(by+c)$$

So you would need to know more about $a,b,c$ before you can say anything. Note that, the argument above assumes $a\neq 0$. Let's take $a=1, b=1, c=0$. Plotting it immediately shows that there are sign changes on either side of the line $x+y=0$. Taking $a=1, b=0, c=0$, however, we can immediately say $f(x,y)=x$ is always positive on the right side of the line $x=0$ and always negative on the left side.

Answer 2

For now if we ignore the $c$ term, then let us understand what $f(x,y)$ in this case means

$$f(x,y) = ax + by = (a,b) .(x,y)$$

Now, $(a,b)$ represents the vector normal to the line $ax+by=0$

Hence, $f(x,y)$ will be positive for any point which has a particular $\cos \theta$ value with this normal, which will have the same sign if and only if they are on the same side of the plane

Answer 3

Points $P$ and $Q$ lie on the same side of the line iff the line segment $\overline{PQ}$ does not intersect it. Define $\sigma:\mathbb R\to\mathbb R^2$, $\sigma:t\mapsto(1-t)P+tQ$. For $0\le t\le1$, $\sigma(t)$ is a parameterization of $\overline{PQ}$, with $\sigma(0)=P$ and $\sigma(1)=Q$. Further, let $g = f\circ\sigma$.

Now, suppose that $P$ and $Q$ lie on the same side of the line and the signs of $f(P)$ and $f(Q)$ differ, i.e., that $g(0)$ and $g(1)$ have different signs. The function $g$ is obviously continuous, therefore by the intermediate value theorem there is some $t_0\in(0,1)$ such that $g(t_0)=0$. This means, however, that the point $\sigma(t_0)\in\overline{PQ}$ lies on the line, a contradiction.