Question: Show that a necessary and sufficient condition for the line $ax+by+c=0,$ where $a,b,c$ are nonzero real numbers, to pass through the first quadrant is either $ac<0$ or $bc<0$.
Solution: We have $ax+by+c=0,$ where $a,b,c$ are nonzero real numbers. This implies that $y=-\frac{a}{b}x-\frac{c}{b}.$
Now since $a,b,c\neq 0$, this implies that the line cannot have equations of the form $y=c_1$ or $x=c_2$ or $y=c_3x,$ $\forall c_1,c_2,c_3 \in\mathbb{R}$.
Therefore the line will pass through the first quadrant only when any one of the three cases described below is satisfied:
Case 1: $-\frac{a}{b}>0$ and $-\frac{c}{b}>0\implies \frac{ac}{b^2}>0\implies ac>0 \text{ (since $b^2>0$)}.$ Now $ac>0$, implies either $a>0$ and $c>0$ or $a<0$ and $c<0$.
Now if $a>0$ and $c>0$, then $-\frac{a}{b}>0\implies b<0 \implies bc<0.$
Again if $a<0$ and $c<0$, then $-\frac{a}{b}>0\implies b>0 \implies bc<0.$
Therefore in any case we have $bc<0$.
Case 2: $-\frac{a}{b}>0$ and $-\frac{c}{b}<0\implies \frac{ac}{b^2}<0\implies ac<0 \text{ (since $b^2>0$)}.$
Case 3: $-\frac{a}{b}<0$ and $-\frac{c}{b}>0\implies \frac{ac}{b^2}<0\implies ac<0 \text{ (since $b^2>0$)}.$
Therefore, after analyzing all the three cases we can conclude that a necessary and sufficient condition for the line $ax+by+c=0,$ where $a,b,c$ are nonzero real numbers, to pass through the first quadrant is either $ac<0$ or $bc<0$.
Is my solution correct and rigorous enough? And, is there a more better and shorter solution?
It is correct but complicated.
As the line, call it (L), does not pass through $0$ (because $c\neq 0$), the given condition is equivalent for (L)
to cross the positive part ($x>0$) of $x$ axis OR
to cross the positive part of ($y>0$) of $y$ axis.
(inclusive "OR" : either one or both axes).
Let us consider the first case : this case is equivalent to say that there exists a point of (L) of the form $(x,y)=(x_0,0)$ with $x_0>0$, i.e. $x_0$ is such that
$$ax_0+b.0+c=0 \iff ax_0=-c$$
which is possible (knowing that $x_0>0$) if and only if $a$ and $c$ have opposite signs, i.e., $ac<0.$
Same reasoning for the other case (dealing with the positive $y$ axis).