Assuming no two lines are parallel, rather than no two lines are not parallel.
Let's label the lines a, b, c, etc. Let's label each intersection as the product of the labels of the intersecting lines. We can then write the intersections as products, then sum them.
2 lines a and b intersect at ab. One intersection (1).
A third line c intersects each of a and b: ac, bc. 3 intersections ab, ac, bc (1+2).
A fourth line d intersects each of the others: ad, bd, cd. 6 intersections ad, bd, cd, ac, bc, ab (1+2+3).
A fifth line e intersects each of the others: ae, be, ce, de. 10 intersections:
ae, be, ce, de, ad, bd, cd, ac, bc, ab (1+2+3+4).
So we have the series: 1, 3, 6, 10, 15, ... generated by an increasing number of lines: 2, 3, 4, 5, 6, ...
So we can write ordered pairs (2,1), (3,3), (4,6), (5,10), (6,15), ... where the first number is the number of lines (L) and the second the number of intersections (I). We can symbolise the pair as (L,I).
If we take L-1, and let n=L-1 we have the series sum 1+2+3+4+...+n which can be rewritten:
(1+n)+(2+n-1)+(3+n-2)+(4+n-3)+... For example: n=6:
1+2+3+4+5+6=(1+6)+(2+5)+(3+4)=7+7+7=(7)(½×6)=21.
For general n then the sum is (n+1)(n/2).
Therefore I=(n+1)(n/2)=(L-1+1)(L-1)/2=½L(L-1). Note that when L=1 there are no intersections. You can check out this formula against the calculations of the ordered pairs already considered.
When L=35, I=½(35)(34)=595 intersections.