# Lines drawn from the intersection of the given lines

Four lines are drawn on a plane with no two parallel and no three concurrent. Lines are drawn joining the points of intersection of these four lines. How many new lines are there now?

I have already found its answer here.

But now I am looking for a more general answer. I mean what if there are $5$ lines or $6$ lines.

Number of points of intersection in case of $5$ lines would be $^5C_2=10.$ Number of lines from these $10$ points=$^{10}C_2=45$. How to proceed from here?

I wish somebody could help.

• What's the problem here? You already have the solution, no? – Parth Thakkar Sep 13 '13 at 18:30

## 2 Answers

The 45 lines include the given ones, and each of the given lines is produced by $\binom 42=6$ different pairs of intersection points so you have to subtract that. This gives $45-5\cdot 6=15$ new lines.

In the general case we can do the same, unless some of the intersection points are unexpectedly collinear (which can, but does not have to, happen when there are at least 6 given lines). But we still get an upper bound for the number of new lines: $$\binom{\binom n2}2 - n\binom{n-1}{2} = \frac{n^4-6n^3+11n^2 -6n}{8}$$ (where I think the polynomial form is correct, but better check my algebra yourself before you use it for anything important, in case I've misplaced some signs along the way).

This upper bound will be reached almost surely if the given lines are chosen randomly with a continuous distribution.

If you have $n$ lines in general position (i.e. no two parallel, no three concurrent) then they will intersect in altogether $\binom n2$ points. Therefore there are $\binom{\binom n2}{2}$ lines through these points. However we have counted a few lines multiple number of times, which we will now have a look at. Note that a line is counted multiple times if it has atleast three points on it. Now this line will be counted as many number of times as there are pairs of points on it. Since each original line intersects every other original line, there are $n-1$ points on every line. Therefore each of the $n$ original lines gets counted exactly $\binom{n-1}{2}$ times. Therefore number of new lines is $\binom{\binom n2}{2} - n\binom{n-1}2$.

P.S.Crosschecked the formula for $n=4$ and it turns out to be correct according to the link provided.