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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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    $\begingroup$ Are you sure that it doesn’t ask for the maximum possible number of additional lines? $\endgroup$ – Brian M. Scott Nov 4 '11 at 22:35
  • $\begingroup$ What if three intersection points lie on a line? Do we call the resulting number of lines 1, 2, or 3? $\endgroup$ – davidlowryduda Nov 4 '11 at 22:38
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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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  • $\begingroup$ The case where three of the intersection points lie on a new line is much more interesting than the original problem! But I'm sure you are right that one was meant to ignore that possibility. $\endgroup$ – Gerry Myerson Nov 5 '11 at 3:36

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