Question

In a plane, there are $37$ straight lines of which $13$ pass through point $A$ and $11$ passes through 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 intersection points the lines have is equal to?

• A. $535$
• B. $601$
• C. $728$
• D. None of these

Answer 1

The correct option is A

$535$

Explanation for the correct option:

Finding the number of points of intersection:

The point of intersection of $37$ lines $={}^{37}C_2$.

However, it is given that $13$ passes through point $A$, so they will have only one intersection point instead of ${}^{13}C_2$ and the $11$ lines that passes through point $B$ will also have only one intersection point instead of ${}^{11}C_2$.

There will be two more intersection points, that are $A$ and $B$.

So, total number of intersection points the lines have is

$$\begin{gathered} ={}^{37}C_2-{}^{13}C_2-{}^{11}C_2+2 \\ =\frac{37\times 36\times 35!}{2\times 35!}-\frac{13\times 12\times 11!}{2\times 11!}-\frac{11\times 10\times 9!}{2\times 9!}+2\left[\because {}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =535 \end{gathered}$$

Hence, option A is correct.