Problem 66 - A Complete Resource Book in Mathematics for JEE Main

Question

Two straight lines intersect at a point \(O\). Points \(A_{1}\), \(A_{2}, \ldots, A_{n}\) are taken on one line and points \(B_{1}, B_{2}, \ldots\) \(B_{n}\) on the other. If the point \(O\) is not to be used, the number of triangles that can be drawn using these points as vertices, is (A) \(n(n-1)\) (B) \(n(n-1)^{2}\) (C) \(n^{2}(n-1)\) (D) \(n^{2}(n-1)^{2}\)

Step-by-Step Solution

Answer

The number of triangles is \(n^2(n-1)\). The correct answer is (C).

Step 1: Understanding the Problem

To solve this problem, we need to determine how many triangles can be formed using the points on two intersecting lines without using the intersection point O. Each triangle must consist of one point from line A and two points from line B, or vice versa.

Step 2: Analyze Selection of Points

Consider the selection of points in the triangle: we need to choose 2 points from one line (either point A's or B's), and 1 point from the other line. This means we have two possible combinations for forming a triangle: selecting 2 points from points \(A_1, A_2, \dots, A_n\) and 1 point from points \(B_1, B_2, \dots, B_n\), and vice versa.

Step 3: Calculate the Combinations

First, calculate the number of ways to choose 2 points from line A (\(A_1, A_2, \dots, A_n\)) and 1 point from line B (\(B_1, B_2, \dots, B_n\)): - Choose 2 points out of \(n\) from line A: \(\binom{n}{2} = \frac{n(n-1)}{2}\) ways.- Choose 1 point from \(n\) points on line B: \(n\) ways.So, the number of triangles in this case is \(\frac{n(n-1)}{2} \times n = \frac{n^2(n-1)}{2}\).

Step 4: Consider the Other Combination

Next, consider choosing 2 points from line B and 1 point from line A:- Choose 2 points out of \(n\) from line B: \(\binom{n}{2} = \frac{n(n-1)}{2}\) ways.- Choose 1 point from \(n\) points on line A: \(n\) ways.So, the number of triangles in this case is \(\frac{n(n-1)}{2} \times n = \frac{n^2(n-1)}{2}\).

Step 5: Sum the Total Number of Triangles

Add up the number of triangles possible from both cases calculated in steps 3 and 4:- Sum is \(\frac{n^2(n-1)}{2} + \frac{n^2(n-1)}{2} = n^2(n-1)\).

Step 6: Conclusion

The number of triangles that can be drawn using these points as vertices, without using the point \(O\), is \(n^2(n-1)\). Therefore, the correct option is (C).

Key Concepts

Triangle Formation from Intersecting Lines

When two lines intersect, they form a central point, known as the intersection point. However, in certain geometry problems, we avoid using this point to construct shapes like triangles. In the given scenario, we're tasked with forming triangles using points on two lines without involving the intersection point, denoted as point O.
To form a triangle, three non-collinear points are needed. The exercise specifies choosing from points on each line. For an effective triangle formation without point O, we can select:

• Two points from the first line and one point from the second, or
• Two points from the second line and one point from the first.

These combinations ensure that none of the selected points lie on the same straight line, thus forming valid triangles.

Role of the Binomial Coefficient

The binomial coefficient is a fundamental concept in combinatorics used to count the number of ways to choose a subset of items from a larger set. In this exercise, the binomial coefficient helps in calculating the number of combinations of points needed to form triangles.
For choosing points from a line, such as selecting two points from line A's set, we use the binomial coefficient denoted as \(\binom{n}{2}\). This is calculated by the formula: \[\binom{n}{2} = \frac{n(n-1)}{2}\]

• This formula gives the number of ways to choose 2 items from n distinct items.
• Similarly, it applies to selecting points from both lines A and B.

In essence, the binomial coefficient simplifies the problem by allowing us to systematically count possible point selections for triangle vertices without manual enumeration.

Understanding Geometry Problems

Geometry problems involving lines and points, like the one tackled here, often require visualizing spatial relationships. By mentally mapping the lines, points, and the intersection, students gain a deeper understanding of how triangles can be formed.
Challenges in geometry problems often stem from:

• Choosing points appropriately to satisfy all conditions (such as avoiding point O).
• Visualizing different possible outcomes and ensuring no cases are overlooked.

Through these exercises, learners can improve their skills in spatial thinking and combinatorial reasoning, which are critical in both theoretical and practical applications of geometry.