Question

There are three coplanar parallel lines. If any p points are taken on each of the lines the maximum no of triangles with vrtices at these points are

Answer 1

hope you will understand this . if yes please give vote .

three coplanar parallel lines means three parallel line which lie in the same plane .

_________________________

_________________________

_________________________

so these are the three coplanar parallel lines.

we have to take P points at each line so,

__.___.___._______......p

___.____._____.____......p

___.___._____._____.......p

like this .

so, we can see that if we want maximum number of triangles , we need no three points in the same line in the three parallel lines . we are only given p coincident points in three parallel lines .

so ,

3p

C

3

this will give us all the no. of 3 points chosen from 3p points but we can see that there are certain points selected which are not forming a triangle.

so, the final equation should be ...

3p p

C - 3. C

3 3

because we need to subtract the points which are collinear in the three parallel lines.

simplifying this we get

3p(3p-1)(3p-2) - 3p(p-1)(p-2) / 6

=> 3p [(3p-1)(3p-2)-(p-1)(p-2)] / 6

=> p/2 × [8p^2 - 6p]

=> p^2 (4p-3)...

hope this answer is correct .

if yes please vote .

Answer 2

Answer:

given p coincident points in three parallel lines

therefore 3p C 3

=> 3p(3p-1)(3p-2) - 3p(p-1)(p-2) / 6

=> 3p [(3p-1)(3p-2)-(p-1)(p-2)] / 6

=> p/2 × [8p^2 - 6p]

=> p^2 (4p-3)

hope this helps!