Question

If n lines are drawn on 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) $\sum\limits_{k=1}^{n-1}{k}$
(b) $n(n-1)$
(c) ${{n}^{2}}$
(d) none of these

Answer 1

Since no 2 lines are parallel, any 2 lines will intersect each other at 1 point. And since no 3 lines are concurrent, no 3 or more lines will intersect at a single point. So we will begin with the 1st line and then will go upto nth line to get the number of different intersection points.

Complete step-by-step answer:
Now we will first draw the first line.

So here on the 1st line we can see that there is no intersection point.
Now we will draw two lines.

So here from the figure we can see that there is 1 intersection point.
Now we will draw three lines.

So here from the figure we can see that there are 2 new intersection points and 1 old intersection point.
Now we will draw 4 lines.

So here from the figure we can see that there are 3 new intersection points and 3 old intersection points.
Hence similarly proceeding for n lines we get,
Number of different points for n lines is new (n-1) points plus old {(n-2)+(n-3)+……….+1}.
So the number of intersection points for n lines is,

& S=(n-1)+(n-2)+(n-3)+.............+1 \\\ & S=\dfrac{n(n-1)}{2} \\\ \end{aligned}$$ **So $$S=\dfrac{n(n-1)}{2}$$ is the same as $$\sum\limits_{k=1}^{n-1}{k}$$. Thus the correct answer is option (a).** **Note:** Knowing the definition of parallel lines and concurrent lines is the key here. Drawing the figure makes the solution clearer. We in a hurry can make a mistake in writing the series for n lines and hence we need to be careful while doing this step. Also we have to expand option (a) to match it with our answer or else we may think option (d) as the answer.