Are there other forms of lines in $\mathbb R \times \mathbb R$? Hilbert's Axioms of incidence for lines are

I1.  For any two distinct point A, B there exists a unique line $l$ containing A, B.
I2. Every line contains at least two points.
I3.  There exist three noncollinear points (that is, three points not all contained in a single line).

We see that the set $\mathbb R \times \mathbb R$ as points and the set of points that satisfy equations of the form $ax+by+c=0$ as lines are a model for the above three axioms.
My question is, is there a different definition of a line in the cartesian coordinates (that is not algebraically equivalent to this definition) that still satisfy the above three axioms?  Or is the form $ax+by+c=0$ fundamentally the only form that can be considered a line? If so, why?
 A: Here's one way to get different "lines" that satisfy these axioms: you can apply any bijective change of coordinates map. 
For instance, apply the coordinate change $x'=x$, $y'=x^2+y$, and then the "lines" of the form $ax' + by' + c = 0$ become $ax + bx^2 + by + c = 0$.
And while the last example was a differentiable change of coordinates with differentiable inverse, one could even apply a discontinuous bijection.

For a different kind of example, one can observe that if $B(\mathcal O,1)$ denotes the open ball of radius $1$ centered at the origin $\mathcal O$, namely 
$$B(\mathcal O,1) = \{(x,y) \in \mathbb R^2 \mid x^2 + y^2 < 1\}
$$
then the set of chords in $B(\mathcal O,1)$ satisfies those axioms; by definition a chord is a nonempty subset of $B(\mathcal O,1)$ which can be expressed as the intersection of $B(\mathcal O,1)$ with some straight line $ax+by+c=0$.
If we now choose a bijection $f : B(\mathcal O,1) \to \mathbb R^2$, then we can define a "line" in $\mathbb R^2$ to be the image under $f$ of a chord, and this set of "lines" will satisfy the axioms. For example one can take 
$$f(x,y) = (\frac{x}{1-x^2-y^2},\frac{y}{1-x^2-y^2})
$$
This example is fundamentally different from the "standard" example, for instance consider a "triangle" in $B(\mathcal O,1)$ whose vertices are on the boundary circle of $B(\mathcal O,1)$ and hence are not points of the geometry. The three sides of that triangle form three chords $L_1,L_2,L_3$ which do not cross each other, but one of the four components of their complement $B(\mathcal O,1) - (L_1 \cup L_2 \cup L_3)$, namely the interior of the triangle, is incident to each of $L_1,L_2,L_3$. This behavior does not happen with ordinary Euclidean lines. 
In fact $B(\mathcal O,1)$ together with its chords form the Beltrami-Klein model of the hyperbolic plane together with its lines. Discovered in the 19th century, this is a model of all of the axioms of Euclidean geometry (hence including Hilbert's incidence axioms) except that it fails to satisfy the parallel postulate.
A: There are many ways to model points as elements in $\mathbb R\times\mathbb R$ without necessarily modelling lines as solutions of linear equations.
Here is a generic way of doing so. Create a bijection $f:\mathbb R\times\mathbb R\to\mathbb R\times\mathbb R$. Any bijection, no particular properties needed. Now, that bijection is going to map "ordinary" points into "ordinary" points, but it may not map "ordinary" lines into "ordinary" lines. Take an example: $f(x,y)=(x,\sqrt[3]{y})$. An ordinary line $ax+by=c$ is mapped by $f$ into the following set of points: $ax+by^3=c$ - those are certainly not straight lines but cubic curves.
Still, by using the fact that $f$ is a bijection and that it maps "incidence" into "incidence", you can conclude that the new model, where points are elements of $\mathbb R\times\mathbb R$ but the lines are curves with equations $ax+by^3=c$ satisfies any incidence axioms that the original model (with "ordinary" lines satisfies), and vice versa.
NB: In other words, you have not only found a non-standard model of Hilbert's axioms, you've found a model isomorphic to the standard one (the isomorphism being the bijection $f$). The question whether there are other, non-isomorphic, models on $\mathbb R\times\mathbb R$ (i.e. is every model on $\mathbb R\times\mathbb R$ of this type with a suitable choice of $f$) would be a separate question, I don't know the answer to it on the top of my head.
