Question

Straight lines are drawn by joining ${}^{\prime }{m}^{\prime }$ points on a straight line to ${}^{\prime }{n}^{\prime }$ points on another line. Then excluding the given points, the number of point of intersections of the lines drawn is (no two lines drawn are parallel and no three lines are concurrent).

• A. $\frac{1}{4}mn(m-1)(n-1)$
• B. $\frac{1}{2}mn(m-1)(n-1)$
• C. $\frac{1}{2}{m}^2{n}^2$
• D. $\frac{1}{4}{m}^2{n}^2$

Answer 1

Answer: (B)

$\frac{1}{2}mn(m-1)(n-1)$

For intersection point we must have two straight lines, for which $2$ points from each straight line must be selected. Now selection of these points can be done in ${}^m{C}_2{\times }^n{C}_2$ ways. Now as shown in diagram these four points can give two different sets of straight lines, which generate two distinct points of intersection.
Then total number of points of intersection is ${}^m{C}_2{\times }^n{C}_2\times 2$.

${}^m{C}_2{\times }^n{C}_2\times 2$.$=\frac{m(m-1)}{2!}\times \frac{n(n-1)}{2!}\times 2$$=\frac{1}{2}mn(m-1)(n-1)$