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Let there be 'p' lines which are concurrent at a point A and 'q' other are concurrent at a point B. Oh, and no 2 lines are parallel.

Case 1: No line out of 'p' passes through B and similarly for 'q' and A. What are the number of points of intersection in the plane?

Case 2: There is one line which passes through both A and B. (the number of lines is still p+q) Now find number of points of intersection.

My doubt is, will the number of points of intersection still be the same?

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  • $\begingroup$ You mean that no line out of A passes through B (in case 1)? $\endgroup$ – Mark Jan 11 '15 at 17:00
  • $\begingroup$ Yes, that is what i mean $\endgroup$ – Saurabh Raje Jan 11 '15 at 17:15
  • $\begingroup$ @SaurabhRaje do you understand what I wrote, are you finding something confusing? $\endgroup$ – Jorge Fernández Hidalgo Jan 11 '15 at 17:26
  • $\begingroup$ I got it. The moment I finished writing, it seemed quite obvious, but my teacher was strongly asserting that the answer shall remain the same, so i had to ask...thanks $\endgroup$ – Saurabh Raje Jan 11 '15 at 17:30
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Suppose there are $a$ lines that pass through $A$ and $b$ lines that pass through $B$ .We shall prove every intersection point except $A$ and $B$ is the intersection point of exactly two lines. You can prove it by contradiction: Suppose there are $3$ concurrent lines, then two of those lines must pass either through $A$ or $B$. but if two of those lines pass through $A$ they already intersected there, so they can't intersect twice. Therefore for every pair of lines, one crossing $A$ and one crossing $B$ there is a unique interection point.

In the other case the answer is $(a-1)(b-1)+2$ since we could just remove that line and we would be left with the previous case with $a-1$ lines thorugh $A$ and $b-1$ lines through $b$. Notice the line passing through $A$ and $B$ didn't add any intersection points.

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