I have been working on exercises in a combinatorics book and have come to a problem that I am having difficulty with. The problem states :
"We are given $n$ straight lines in the plane, in general position, i.e., s.t. no two of them are parallel and no three of them pass through the same point. Compute the number of regions into which these lines divide the plane."
In the hints it says that there are $\binom{n}{2}$ distinct intersection points for the lines. They then talk about choosing a line $r$ which is not parallel to any straight line passing through two of these $\binom{n}{2}$ points. There are then two kinds of regions : those which possess a highest point with respect to $r$ -- which is necessarily one of the $\binom{n}{2}$ points of intersection of the $n$ given lines-- and those which don't possess a highest point. They say that one step of the solution is to show that there is a bijection between the regions of the first kind and the set formed by the $\binom{n}{2}$ points. The second step is to show that there are exactly $n+1$ regions of the second kind, so that the total number of regions is equal to $\frac{n(n-1)}{2} + (n+1)$.
Can someone show me the to construct the bijection referred to in the hint and how to determine that there are $n+1$ regions of the second kind ?
