I want to use the Szemerédi–Trotter theorem (for all m, n $\leq$ 1, we have that the number of point-line incidences in a set of n points and m lines in $\mathbb{R}$$^2$ is at most O(m$^{2/3}$n$^{2/3}$ + m + n)) to show that, given an n-point set P in the plane, the number of distinct lines containing at least k points of P is at most O(n$^2$/k$^3$ + n/k), and the number of point-line incidences determined by such lines is at most O(n$^2$/k$^2$ + n).
Does anyone have a proof?
The way to read the big-oh in Szemerédi–Trotter is really that the number of point-line incidences (call that number $p$) is at most $\max\{O(m^{2/3}n^{2/3}), O(m), O(n)\}$. In other words, at least one of the following holds: $$p = O(m^{2/3} n^{2/3}) \text{ or } p = O(m) \text{ or } p = O(n).$$
If we apply Szemerédi–Trotter to the system of all $n$ points and the $m$ lines which pass through at least $k$ of them, we have $p \ge km$, since each line contributes at least $k$ point-line incidences. Therefore we have $$ km = O(m^{2/3} n^{2/3}) \text{ or } km = O(m) \text{ or } km = O(n)$$ which tells us that $$ m = O(n^2/k^3) \text{ or } k = O(1) \text{ or } m = O(n/k).$$
We can rule out the case that $k = O(1)$ because then both of your conclusions hold trivially. In the other cases, Szemerédi–Trotter tells us (bearing in mind which case we're in) that $$m = O(n^2/k^3) \implies p = O(n^2/k^2)\quad \text{ or } \quad m = O(n/k) \implies p = O(n).$$
Finally we can combine the two bounds on $m$ and $p$ into $m = O(n^2/k^3 + n/k)$ and $p = O(n^2/k^2 + n)$ to write them more concisely.