Let $\mathcal{P}$ be a projective plane (not necessarily desarguesian) and $\mathcal{L}$ its set of lines. Then any $d_1,d_2 \in \mathcal{L}$ are in bijection (i.e. they have the same cardinal). When this cardinal is a finite number $q+1$ ($q$ is called the order of $\mathcal{P}$), then it is well known that both $\mathcal{P}$ and $\mathcal{L}$ have $q^2+q+1$ elements. So one has a non canonical bijective map between $\mathcal{P}$ and $\mathcal{L}$.
Now for my question: Is there a more conceptual way to prove this bijection, especially in the case of planes of infinte order?
All the proofs I'm aware of work by computing the cardinal of each set and finding that they are equal, and obviously this works only for finite orders.
For any line $l\in\mathcal{L}$ there are distinct points $p_1,p_2\in\mathcal{P}$ such that $l=\overline{p_1p_2}$, so the map $$\mathcal{P}^2-\Delta(\mathcal{P})\ \longrightarrow\ \mathcal{L}:\ (p_1,p_2)\ \longmapsto\ \overline{p_1p_2},$$ is surjective, where $\Delta(\mathcal{P})\subset\mathcal{P}^2$ denotes the diagonal. Similarly, for any point $p\in\mathcal{P}$ there exist distinct lines $l_1,l_2\in\mathcal{L}$ such that $p=l_1\cap l_2$, so the map $$\mathcal{L}^2-\Delta(\mathcal{L})\ \longrightarrow\ \mathcal{P}:\ (l_1,l_2)\ \longmapsto\ l_1\cap l_2,$$ is also surjective, where $\Delta(\mathcal{L})\subset\mathcal{L}$ again denotes the diagonal. This shows that $$|\mathcal{P}|\times(|\mathcal{P}|-1)\geq|\mathcal{L}|\qquad\text{ and }\qquad|\mathcal{L}|\times(|\mathcal{L}|-1)\geq|\mathcal{P}|$$ Of course when $|\mathcal{P}|$ is infinite then $$|\mathcal{P}|=|\mathcal{P}|\times(|\mathcal{P}|-1)\qquad\text{ and }\qquad|\mathcal{L}|=|\mathcal{L}|\times(|\mathcal{L}|-1),$$ from which it follows that $|\mathcal{P}|\geq|\mathcal{L}|\geq|\mathcal{P}|$, so the two are equal.