Q: There are n straight lines in a plane, no two of which are parallel and no three of which pass through the same point. How many additional lines can be generated by means of points of intersections of the given lines.

Sol. Let AB be any one of the n straight lines and let it be intersected by some other line CD at point E.

Note that line AB contains (n – 1) such points. It follows that there are n (n – 1) such points of intersection. Since each of the points of intersection occurs in two of the n lines, the number of such points is $\displaystyle \frac{1}{2}n(n-1)$

We shall now find the number of additional lines passing through these points. The number of additional lines passing through points of the type E is equal to the number of points lying outside the line AB and CD, because a new line is obtained through E only if the other point lies outside AB and CD. Since AB and CD each contain (n – 2) new points, the number of new points outside AB and CD is

$\displaystyle \frac{1}{2}n(n-1) – [(n-2) + (n-2) + 1]$

$\displaystyle = \frac{1}{2}n(n-1) – (2n-3)$

Thus, the number of lines through E is $\displaystyle \frac{1}{2}n(n-1) – (2n-3)$ .

Since there are $\displaystyle \frac{1}{2}n(n-1)$ new points, the number of new lines is $\displaystyle \frac{1}{2}n(n-1) [\frac{1}{2}n(n-1) – (2n-3)] $

But in this way each line is counted twice. Therefore, the required number of lines is

$\displaystyle = \frac{1}{2} .\frac{1}{2}n(n-1) [\frac{1}{2}n(n-1) – (2n-3)] $

$\displaystyle = \frac{1}{8}n(n-1) (n-2) (n-3) $