Question

$n$ lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. The number of different points at which these lines will cut is

• A. $\sum _{k=1}^{n-1}k$
• B. $n(n-1)$
• C. $n^2$
• D. None of these

Answer 1

The correct option is A $\sum _{k=1}^{n-1}k$

Since according to question no two lines are parallel and no 3 of them are concurrent;this means any two selected line intersects at a point not shared by any other intersection point.
Hence no. of total intersection points =$\left(\begin{array}{c}n\\ 2\end{array}\right)\times 1=\frac{n(n-1)}{2}$
which is same as option A since ${\sum }_{k=1}^{n-1}k=\frac{(n-1)n}{2}$