Some point in a line $Ax+By+C=0$ Given a line in a 2D space, defined by equation $Ax+By+C=0$, is it possible to find a point of it without assume that A or B are different of 0 ?
That is, usually the method to find some point is "assume $A \ne 0$, if we fix $y=0$ the solution of the equation gives that point $(-C/A,0)$ is a point of the line, otherwise $B \ne 0$ and ...·
But it is possible any other method without split the problem in two ?
In other words, is it possible to find an expression for some point (any one) of the line that doesn't contains a division by A, B or C or by any other term that can be zero in some cases ?
The question could be expressed in another way: given a line $Ax+By+C=0$, give an expression of the same line in vector/parametric form that is valid for any value of A, B, C. 
The direction vector is easy to find, (-B,A), the remainder target is to find the expression of some point.
 A: Consider the additional line $Bx-Ay=0$. The linear system
\begin{cases}
Ax+By=-C\\[4px]
Bx-Ay=0
\end{cases}
has solution
$$
x=-\frac{AC}{A^2+B^2}
\qquad
y=-\frac{BC}{A^2+B^2}
$$
Comments
The additional line is the perpendicular passing through the origin and we found the intersection of the two lines. If $C=0$, we of course get $(0,0)$, but no assumption on $C$ is actually necessary.
Since either $A$ or $B$ is nonzero, we have $A^2+B^2\ne0$, so the division doesn't pose problems.
A graphic example with $A=3$, $B=2$, $C=5$ that shows we're essentially finding the point having minimal distance from the origin.

If instead you consider as additional line $Bx-Ay=t$, for a variable $t$, the solution is
$$
x=-\frac{AC-Bt}{A^2+B^2}
\qquad
y=-\frac{BC+At}{A^2+B^2}
$$
and, as $t$ varies, you get all points on the given line.
A: How about
$$( -\frac{AC}{A^2 + B^2}, - \frac{BC}{A^2 + B^2})$$
A: For $C=0$, we can choose


*

*$x=-B$

*$y=A$


For $C\neq0$ and $A=1$, we can choose


*

*$x=-C-By$


For $C\neq0$ and $B=1$, we can choose


*

*$y=-C-Ax$


For $C\neq0$ and $C=kA$, we can choose


*

*$x=-B-k$

*$y=A$


For $C\neq0$ and $C=kB$, we can choose


*

*$x=-B$

*$y=A-k$


otherwise we need always division.
A: This is an interesting question. I think the answer is "no".
If $C = 0$ then the point 
$$
\left( \frac{AB^2}{A^2+B^2},  \frac{-A^2 B}{A^2+B^2} \right)
$$
will do, but if $A=0$ and $C \ne 0$ then $y$ must be $-C/B$. You have to divide by $B$. Any general formula you propose to deal with the general problem will have a $B$ in the denominator.
Edit. As @gimusi  points out, $(B, -A)$ works when $C=0$. My solution was clumsy.
Edit. My "think so" is wrong. Other answers are better. At least my use of $A^2 + B^2$ was on the right track. It's the determinant of the matrix in @egreg 's solution.
