Let $P$ denote the set of intersection points, and $\deg{p}$ denote the number of lines that go through point $p \in P.$ $\binom{N}{2}$ counts the number of pairs of lines. $\sum_{n\geq2} a_n \binom{n}{2}$ counts the number of pairs of parallel lines. We show that $\sum_{n\geq2} b_n \binom{n}{2}$ counts the number of pairs of intersecting lines. Keep a counter at each intersection point. For each pair of nonparallel lines, add one to the counter at the point of intersection of these two lines. Then the sum of all intersection points' counters is the number of pairs of intersecting lines. By double counting, this is also just $\sum_{p \in P} \binom{\deg{p}}{2},$ because $\binom{\deg{p}}{2}$ will be the value of the counter at point $p$ after all pairs of lines that intersect at $p$ are counted for. By grouping points with the same degree - the number of points with degree $m$ is $b_m$ by definition - we get that this sum is equal to $\sum_{n\geq2} b_n \binom{n}{2},$ so we are done.