Question

there are 8 points lying on a plane such that no two of the lines joining them are parallel and no three are concurrent except at the given points. If all the possible lines are drawn passing through every pair of these points what is the number of distinct points of intersection of the lines excluding the original 8 points?

Answer 1

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Answer 2

Given:

8 point is on the plane in which no two lines are parallel and no three lines are concurrent.

To find:

Number of total distinct points of intersection.

Solution:

Suppose their are n number of points lying in a same plane which have no two lines are parallel and no three lines are concurrent.

This means that any selected two lines can intersect with each other at a point and not shared by any other intersection point.

Therefore, number of total intersection points are

= $\binom{n}{2} \times 1 = \frac{n(n-1)}{2}$

So, according to question there are 8 points lying in a plane

therefore, n = 8

by substituting the value of n in above equation we get,

number of total intersection points = $\frac{n(n-1)}{2}$

                                                          = $\frac{8(8-1)}{2}$

                                                          = $\frac{8 \times 7}{2}$

                                                          = 27

Hence, total number of intersection points are 27.