Question

In a plane there are $37$ straight lines, of which $13$ pass through the point $A$ and $11$ pass through the point $B$. Besides, no three lines pass through one point, no line passes through both points $A$ and $B$,and no two are parallel. Then the number of inter-section points the lines have is equal to:

• A. $535$
• B. $601$
• C. $728$
• D. $548$

Answer 1

The correct option is A $535$

So if we have $n$ lines and if no three lines passes from one point and none of them are parallel then total number of points of intersections are given by ${}^n{C}_2$ so here we have total of $37$ lines so ideally we should have ${}^{37}{C}_2$ points

But it has been mentioned that $13$ passes from a point and $11$ passes from the other point so if you will choose a $2$ lines from these lines, then they are going to give same point again and again so we have to remove these points i.e now we have ${}^{37}{C}_2$ - ${}^{13}{C}_2$ - ${}^{11}{C}_2$ $+2$

That $2$ is because we will get two points, one is $A$ and other is $B$ so we have to add them back.

And this give a total of $535$ points.