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 point 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

Hint: In this question, we need to find out the number of intersection points from the given particular.
A point of intersection means the point in which two lines intersect. We need to consider all the given particulars and find out the total possible point of intersection. Then subtracting the points, we have calculated extra we will get the required result.
Formula to be used:
\[C(n,r){ = ^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]

Complete answer:
We have, in the plane there are \[37\] straight lines.
In which \[13\] pass through point A and \[11\] pass-through point B. Also, no three lines pass through one point, no line passes through both A and B and no two are parallel.
We need to find out the number of intersection points the lines have.
A point of intersection means the point in which two lines intersect.
In the plane, there are \[37\] straight lines that have \[^{37}{C_2}\] points of intersection since we need two straight lines to create a point of intersection and the total numbers are thirty-seven.
\[13\] straight lines pass through point A yield only one intersection point instead of \[^{13}{C_2}\] and \[11\] straight lines pass through point B also yield only one intersection point instead \[^{11}{C_2}\].
Therefore, the number of total intersection points =\[^{37}{C_2}{ - ^{13}}{C_2}{ - ^{11}}{C_2} + 1 + 1\]
=\[\dfrac{{37!}}{{2!(37 - 2)!}} - \dfrac{{13!}}{{2!(13 - 2)!}} - \dfrac{{11!}}{{2!(11 - 2)!}} + 1 + 1\]
\[ = \dfrac{{37!}}{{2!35!}} - \dfrac{{13!}}{{2!11!}} - \dfrac{{11!}}{{2!9!}} + 2\]
\[ = \dfrac{{37 \times 36 \times 35!}}{{2 \times 1 \times 35!}} - \dfrac{{13 \times 12 \times 11!}}{{2 \times 1 \times 11!}} - \dfrac{{11 \times 10 \times 9!}}{{2 \times 1 \times 9!}} + 2\]
\[ = 37 \times 18 - 13 \times 6 - 11 \times 5 + 2\]
\[ = 666 - 78 - 55 + 2\]
\[ = 535\]
Hence, the number of intersection points equals \[535\].

Therefore, option (A) is the correct option.

Note:
Combination:
The combination is represented as a selection of items from a collection, such that the order of selection does not matter.
For a combination,
\[C(n,r){ = ^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
Where, the factorial of n is denoted by n! and defined by,
\[n! = n(n - 1)(n - 2)....2 \times 1\]