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  • There are three coplanar parallel lines. If any  points are taken on each the lines, the maximum number of triangles with vert

There are three coplanar parallel lines. If any  p points are taken on each the lines, the maximum number of triangles with vertices at these points is

Option: 1

3p^{2}(p-1)+1


Option: 2

3p^{2}(p-1)


Option: 3

p^{2}(4p-3)


Option: 4

none\:of\:these


Answers (1)

best_answer

 

Rule for Geometrical Permutations -

There are n points in a plane out of these points no three are in the same line except m points which are collinear, then

(1) Total number of different lines obtained by joining these n  points is nC2 - mC2 + 1

(2) Total number of different trangles obtained by joining these n  points is nC2 - mC2 

-

 

 

Maximum no. of triangle

=\frac{3p(3p-1)(3p-2)-3p(p-1)(p-2)}{6}

=\frac{p}{2}\left [ 9p^{2}-9p+2-p^{2}+3p-2 \right ]=p\left [ 4p^{2}-3p \right ]

=p^{2}\left [ 4p-3 \right ]

Posted by

Kuldeep Maurya

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