some points are plotted on a plane such that any three of them are noncollinear.each point is joined with all remaining points by line segments.find the number of points if the number of line segments are 10
Answers
Take any 10 points in a plane and name them as A, B ,C ,D ,E ,F ,G ,H ,I ,J.
Now next statement says 4 of these points are joined to 6 of the remaining points. So divide these 10 points into two groups containing 4 and 6 points arbitrarily. Let's assume first group as A, B, C, D and second group contains remaining 6 points. Now each point in first group is joined to each point in second group.
So line segments formed are 6+6+6+6=6*4=24. Because A is joined to 6 points, B is joined to 6 points and so on.
Next statement says each point from the group of 6 is joined to 5 other points (these 5 points can be in any group). Now these 6 points are already connected to 4 points in other group, so we need to connect each point to 1 more point from the same group. So divide these group of six into 3 groups of 2 and each group of 2 will form one line segment. For example E and F, G and H, I and J. Here if any point is repeated in group of two, then it will be joined to more than 5 points so we don't repeat. Now the given condition is satisfied.
So answer is 24+3=27.
We know that number of sides in a quadrilateral are 'n'.
Also number of diagonals in a quadrilateral are '(n(n-3)) /2'. So number of line segments in a quadrilateral = no.of sides + no.of diagonals
Therefore n+(n(n-3))/2 = 10, 2n+(n(n-3))=20 ,
2n+n^2 - 3n=20,
n^2 - n=20, n^2 - n - 20 = 0,
n^2 - 5n +4n - 20=0, n(n - 5)+4(n - 5)=0,
(n - 5)(n + 4)=0,
Therefore n = 5 or n = -4. Number of points cannot exist in negative form,So number of points are '5'