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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?

$4C_{2}=6$

$6C_{2}=15 -2 -2 -2=7$

New lines obtained = $7-4$ = $3$

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    $\begingroup$ Where is the question? $\endgroup$ – Marc van Leeuwen Mar 5 '13 at 6:00
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    $\begingroup$ Rajesh: It would be good for you to consider accepting answers when they are helpful. You can accept one answer per question, and you can do so by clicking on the $\checkmark$ to the left of the answer you'd like to accept. Plus, bonus: you get two reputation points each time you accept an answer! $\endgroup$ – Namaste Mar 7 '13 at 9:08
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You start with four lines and six points of intersection. If you draw the figure, there are three of the six points on each line, so the six points do not determine fifteen lines. You don't explain the reasoning behind the three subtractions of two. Presumably you are claiming that the end figure has seven lines in it. I believe that is true, but it needs some argument. I would say that each point has two lines through it which connect it to four other points, so you can construct one new line through each point, but have counted each new line twice, giving three.

A figure is below. The light lines are the original four, the dots are the points, and the three heavy lines are the new ones.enter image description here

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  • $\begingroup$ If possible, can you draw the diagram please? In my diagram I am getting 4 lines. $\endgroup$ – Ramit Sep 12 '13 at 12:55
  • $\begingroup$ @Ramit: I have added a figure. I don't see where the fourth line goes. $\endgroup$ – Ross Millikan Sep 12 '13 at 16:47
  • $\begingroup$ I was actually counting the wrong lines. Thanks a lot for your help. $\endgroup$ – Ramit Sep 12 '13 at 17:46
  • $\begingroup$ If you don't mind, may I ask you to please elaborate further. I mean you are talking about three subtractions of two as done by OP. But I am failing to understand the reasoning behind these subtractions. Moreover, what is $^6C_2$ for? $\endgroup$ – Ramit Sep 13 '13 at 10:43
  • $\begingroup$ @Ramit: I don't understand OP's calculation, either. $^6C_2$ is the number of ways to choose two of the six points, which would be the total number of lines if no three points were collinear. I think the three $-2$s were to account for the ones missing because of collinearity, in which case there should be $4$ of them and the result $7$ is correct. $\endgroup$ – Ross Millikan Sep 13 '13 at 13:15

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