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A plane has 6 lines of which no two lines are parallel and no three are concurrent. Their points of intersection are joined, how many of additional lines are so formed?

I know that number of points of intersection for $n$ lines would be $\sum \limits_{i=1}^{n-1} i=\frac{n(n-1)}{2}$, but then how do I do the rest?

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    What if three intersection points lie on a line? Do we call the resulting number of lines 1, 2, or $3$?2011-11-04

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Every choice of two intersection points determines a line. You already have some of these -- how many pairs of intersection points will generate each of the original 6 lines?

There's a risk that three of the intersection points will lie on a common line that was not one of the originals, but you're probably supposed to ignore that possibility.

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    The case where three of the i$n$tersection points lie on a ne$w$ line is much more interesting than the original problem! $B$ut I'm sure $y$ou are right that one was meant to ignore that $p$ossibility.2011-11-05