If there are $n$ points in a plane and no three of them are collinear, then find Total no of intersection of the lines joining these $n$ points.

combinatorial-geometrycombinatoricspermutations

Question:- If there are $n$ points in a plane and no three of them are collinear, then find Total no of intersection of the lines joining these $n$ points.

Since no three points are collinear, then the total no of lines formed by these $n$ points is $\binom {n}{2}$. If we select any two lines then we will get an intersection point, So total no of intersections of these $p=\binom {n}{2}$ lines is $\binom {p}{2}$.This also matches with the answer in the book.

But Later on I realise that there should be two restrictions while applying this formula

– When Some of the lines are parallel

Suppose we have four Non- collinear points in a plane namely $A(5,5)$, $B(1,2)$, $C(2,-2)$, $D(10,4)$

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Note that Line $AB$ and $CD$ are parallel

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In the formula $\binom {p}{2}$ which gives no of intersection of lines, we have taken intersection point of these two lines in count as well but since they are parallel, they have no intersection point,so the no of points intersection will be one less than the value obtained by the above formula.

In the general case many of the lines can be parallel, so we should impose the restriction that no two lines should be parallel.

– When Lines are concurrent(They have common intersection Point)

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When some of the lines are concurrent, then they have same intersection point but In the formula $\binom{p}{2}$ those points are treated as different so we have to subtract some number in order correct overcounting the no of cases.

So we should have restriction that no three lines are concurrent.

  • Are my restrictions correct.
  • Are there any other resctrictions.

Best Answer

Yes, you are correct; the $p\choose2$ value is merely an upper bound.

For it to be attained, you need there to be $p$ distinct lines, each pair of which intersects in a unique point. That the lines are distinct is to enforce that no three points are colinear. That each pair intersects somewhere is to enforce that lines are not parallel. That the intersections are unique is to enforce that no three lines are concurrent. If you've imposed these three restrictions, then the quoted value will be correct. (You also need the points to be distinct for the lines to be well-defined in the first place, but this is a reasonable implicit assumption to make in the problem statement.)

One might describe point/line arrangements like this as being in "general position", though the term's meaning is flexible depending on context.

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