Arrange 10 points on five lines where each line(intersecting) has exactly 4 points One possible case is that forming a star and then arranging 10 points on its vertices.

Is there any other possible case for this arrangement? If not then how can we prove it mathematically?
Degenerated cases like concurrent lines and parallel lines are also allowed.
 A: Simply draw your lines in such a way they all intersect in distinct points. This will give rise to $\binom{5}{2}=10$ intersections and since each line intersects the other 4 in distinct locations, each line will have 4 intersections. Place your points on these intersections and you will have your solution. 
Since $\binom{5}{2}=10$ is the maximum value for $\binom{5}{k}$ for all $0\le k\le 5$, we know we can't do any better.
EDIT: The key to maximizing these incidences is to share the point with as many lines as possible, thus maximizing the number of incidences per point. Since a system of $n$ lines can intersect in at most $\binom{n}{2}$ points, I can't increase this number of intersections by allowing three or more lines to intersect at a single point. And even if I do allow this, there is a one-for-one trade in intersections versus incidences since the number of points each line contains is fixed.
BTW have you heard of the Szemeredi-Trotter theorem? It isn't very helpful here (the size of the system is too small), but it gives an almost tight asymptotic bound on the number of incidences a system can have.
