Question

Straight lines are drawn by joining m points on a straight line to n points on another line. Then excluding the given points, the number of point of intersection of the lines drawn is (no two lines drawn are parallel and no three lines are concurrent)
(a) $\dfrac{1}{4}mn\left( m-1 \right)\left( n-1 \right)$
(b) $\dfrac{1}{2}mn\left( m-1 \right)\left( n-1 \right)$
(c) $\dfrac{1}{2}{{m}^{2}}{{n}^{2}}$
(d) $\dfrac{1}{4}{{m}^{2}}{{n}^{2}}$

Answer 1

Hint: At first, choose points A, B from lines ${{l}_{1}}$ and ${{l}_{2}}$ containing m and n points each which can be done in ${}^{m}{{C}_{1}}$ , ${}^{n}{{C}_{1}}$ ways respectively. Now we are left with $\left( m-1 \right)$ , $\left( n-1 \right)$ so to choose point C, D it can be done in ${}^{m-1}{{C}_{1}},{}^{n-1}{{C}_{1}}$ ways. So, total number of ways is ${}^{m}{{C}_{1}}\times {}^{n}{{C}_{1}}\times {}^{m-1}{{C}_{1}}\times {}^{n-1}{{C}_{1}}$. We observe that C, D can be chosen first and the A and B but the work will remain the same so the ways will be the total number of ways divided by 2.

Complete step-by-step answer:
In the question, straight lines are drawn by joining m points on a straight line to n points on another line. Then excluding the points given, we have to find the total number of points of intersections of the lines drawn given the condition that no two lines drawn parallel and no three straight lines are concurrent.

Let us consider two straight lines ${{l}_{1}}$ , ${{l}_{2}}$ . ${{l}_{1}}$ containing m points and ${{l}_{2}}$ containing n points.
So, let’s suppose we choose 1 point from ${{l}_{1}}$ and another point from ${{l}_{2}}$ named as A, B respectively and join it to form AB line.
We can choose point A in ${}^{m}{{C}_{1}}$ ways and point B in ${}^{n}{{C}_{1}}$ ways.
Now similarly, we will choose 1 more point ${{l}_{1}}$ and again another one from ${{l}_{2}}$ named C, D respectively and join it to form a CD line.

After choosing A, B we were left with $\left( m-1 \right)$ and $\left( n-1 \right)$ points on the line so, the ways of choosing point C on line ${{l}_{1}}$ is ${}^{m-1}{{C}_{1}}$ and choosing D from ${{l}_{2}}$ is ${}^{n-1}{{C}_{1}}$ .
We now see that for one intersection line A B and C D has been used.
We can also see that if point C, D chose fist and the A, B then also the work completed will be the same. So, the same is being completed in two ways.
The total number of ways to complete the work is ${}^{m}{{C}_{1}}\times {}^{n}{{C}_{1}}\times {}^{m-1}{{C}_{1}}\times {}^{n-1}{{C}_{1}}$ .

As we know that work was repeated so, it will be divided by 2 which means the total number of points of intersection will be:
$\dfrac{1}{2}\times {}^{m}{{C}_{1}}\times {}^{n}{{C}_{1}}\times {}^{m-1}{{C}_{1}}\times {}^{n-1}{{C}_{1}}$
$\Rightarrow \dfrac{mn\left( m-1 \right)\left( n-1 \right)}{2}$
Hence, the correct option is ‘b’.

Note: We can solve the problem by finding the number of quadrilaterals made by 2 points of the 1st line and 2 points of the 2nd line. Now, on visualizing we observe that there are two points of intersection, one is when diagonals intersect and other is when two lines which we actually drew to make the quadrilaterals intersect hence, the number of intersections will be number of quadrilaterals $\times$ 2 which is ${}^{m}{{C}_{2}}\times {}^{n}{{C}_{2}}\times 2$ .