Prove: In a finite geometry, if a line has $n$ points, then the total number of points is $n^2 - n + 1$. Prove: In a finite geometry, if a line has $n$ points, then the total number of points is $n^2 - n + 1$. The following axioms define a finite geometry:
1) Any two distinct points are a unique line.
2) If p and q are distinct lines then they is a unique point on both.
3) There are at least 3 distinct points on any line.
4) Not all the points are on the same line. 

Assume there exists a line $l$ with $n$ points: $p_1,...p_n$. From 4) there exists $p'$ in the geometry and not on the line. With the pairs $(pi,p')$, I can form $n$ lines. I know that each of these lines has at least 3 points, but I don't see why they can't be one of the $n_j$. I might be missing something, but I can't seem to figure out how to use 2) and 3) to get more points.
Any help would be welcome!
 A: The stock proof of this is the one you started with. 
Let $\ell $ be any line, with points $P_1, \ldots, P_n$ (where $n$ is at least $3$). Let $Q$ be any point not on $\ell$ (axiom 4). Consider the lines 
$$
\ell_i (i = 1, \ldots, n)
$$
where $\ell_i$ contains both $Q$ and $P_i$. These are all distinct (axiom 1), and they intersect pairwise exactly at the point $Q$ (axiom 2). 
Claim: every point of the projective plane lies on one of the $\ell_i$. Suppose point $A$ does not. Then the line $QA$ ($Q$ and $A$ are distinct, because $Q$ lies on all the $\ell_i$, and $A$ lies on none of them!) is parallel to $\ell$, which is impossible by axiom 2.
Claim: for any distinct $i, j$, $\ell_i$ and $\ell_j$ have the same number of points. 
Proof: let $P_k$ be different from $P_i$ and $P_j$. Then $P_k$ lies on neither $\ell_i$ not $\ell_j$ (why not? You need to verify this!). 
Let the points of $\ell_i$ be $A_1 \ldots A_p$; consider the lines $m_i$ from $P_k$ through $A_i$. The line $m_i$ intersects $\ell_j$ in some point $B_i$. And this provides a bijection from the $A$s to the $Bs$. 
Claim: the number of points on $\ell_1$ is the same as the number $n$ of points on $\ell$. 
Reason: Consider the line $\ell_2$. It contains the points $Q$ and $P_2$, and must contain another point $S$ distinct from these, by axiom 4. $S$ is not a point of $\ell$, for then $\ell$ and $\ell_2$ would both contain $P_2$ and $S$, and hence be the same (axiom 2). Nor, for similar reasons, is $S$ a point of $\ell_1$. Now for each point $P_i$ of $\ell$, define a point $R_i$: $R_i$ the the intersection of the line $SP_i$ with $\ell_1$. This defines a bijection from $\ell$ to $\ell_1$, hence they have the same number of points. 
Claim: the total number of points is $n^2 - n + 1$. For there are $n$ lines $\ell_i$, each with $n$ points, for a total of $n^2$ points. But the point $Q$ lies on all of them, which means it's been overcounted $n-1$ times. QED. 
