Every line in a finite projective plane has the same number of points

Let $\mathrm{\ell }$ and $m$ be distinct lines, and write $p:=\mathrm{\ell }\cap m$, which exists and is unique by [L1].

There is a point $x$ outside $\mathrm{\ell }\cup m$. If one of the four noncollinear points from [L1] lies outside $\mathrm{\ell }\cup m$, choose it. Otherwise exactly two of them lie on $\mathrm{\ell }$ and two lie on $m$; let $a\in \mathrm{\ell }\setminus m$ and $c\in m\setminus \mathrm{\ell }$ be such points. The line through $a$ and $c$ is distinct from both $\mathrm{\ell }$ and $m$, so because every line contains at least three points, it has a point $x$ different from $a$ and $c$, and that $x$ lies on neither $\mathrm{\ell }$ nor $m$.

For each $y\in \mathrm{\ell }\setminus \{p\}$, let $\phi (y)$ be the unique point where the line through $x$ and $y$ meets $m$. Since $x\mathrm{\notin }m$, one has $\phi (y)\mathrm{\ne }p$, and if $\phi (y_1)=\phi (y_2)$ then the lines through $x$ and $y_1,y_2$ coincide, forcing $y_1=y_2$. So $\phi$ is injective from $\mathrm{\ell }\setminus \{p\}$ to $m\setminus \{p\}$.

Reversing the same construction with the roles of $\mathrm{\ell }$ and $m$ exchanged gives an injective map from $m\setminus \{p\}$ to $\mathrm{\ell }\setminus \{p\}$. Since the two sets are finite, they have the same cardinality. Therefore $\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }=\mathrm{\mid }m\mathrm{\mid }$.