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# Intersection of n lines

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• ### Intersection of n lines

1. 1. 21
2. 2. 28
3. 3. 36
4. 4. 45
5. 5. 55
6. 6. 66
7. 7. 78
8. 8. I 91
9. 9. II e e I 105e
10. 10. Derivation• Two lines cut in 1 point.• A third line will cut the other two lines in 2 more points, giving 1 + 2 = 3 points.• A fourth line will cut the other 3 lines in 3 more points, giving 3 + 3 = 6 points.• So the series goes n = 1, 2, 3, 4, 5, 6, .......Number of points 0, 1, 3, 6, 10, 15, .......The gap increases by 1 each time.• This is the sequence of triangular numbers which has the nth term given by n(n-1)/2. OR• If you made up a difference table, the second differences would be constant, so the nth term is a quadratic in n.• If you assume f(n) = an^2 + bn + c, where f(n) is the nth term n = 1 a + b + c = 0 n = 2 4a + 2b + c = 1 n = 3 9a + 3b + c = 3 and solving these 3 equations for a, b, c, a = 1/2, b = -1/2, c = 0• So f (n) = n^2/2 - n/2 = n(n-1)/2 as given above. So with n lines there are n (n-1)/2 intersection points.