Line L1 of slope 2 and line L2 of slope 1/2 intersect at the origin O. In the first quadrant, P1, P2

Q: Wait, why do we subtract $^{13}C_{3}$ and $^{10}C_{3}$? Aren't we double-counting the origin?

That is a brilliant observation! We are not double-counting. We are using the principle of inclusion-exclusion for disjoint sets of collinear points. The origin $O$ is part of the $13$ points on $L_1$ and the $10$ points on $L_2$. When we subtract $^{13}C_{3}$, we remove all combinations of $3$ points that lie on $L_1$. When we subtract $^{10}C_{3}$, we remove all combinations of $3$ points that lie on $L_2$. Since a set of $3$ points cannot be collinear on both $L_1$ and $L_2$ simultaneously (unless they are all the origin, which is impossible for $3$ distinct points), these two sets of invalid triangles are mutually exclusive. Thus, simple subtraction is correct.

Q: Can we just pick one point from $L_1$, one from $L_2$, and the origin to form a triangle?

Yes, absolutely! That is one of the many valid ways to form a triangle. The beauty of the 'Total minus Invalid' method is that it automatically accounts for all valid triangles, including those that use the origin as a vertex and those that do not. You don't need to manually count every case; the combinatorics does the heavy lifting for you.

Q: Why is the total number of points $22$? I thought it was $12 + 9 = 21$.

Don't forget the origin! The problem states $P_1, \dots, P_{12}$ are on $L_1$ and $Q_1, \dots, Q_9$ are on $L_2$, and both lines intersect at the origin $O$. So, the set of points is $\{O, P_1, \dots, P_{12}, Q_1, \dots, Q_9\}$. That is $1 + 12 + 9 = 22$ points. Always count your vertices carefully before starting the combination formula!

Question

Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin O. In the first quadrant, $P_1, P_2, \dots, P_{12}$ are 12 points on line $L_1$ and $Q_1, Q_2, \dots, Q_9$ are 9 points on line $L_2$. Then the total number of triangles, that can be formed having vertices at three of the 22 points $O, P_1, P_2, \dots, P_{12}, Q_1, Q_2, \dots, Q_9$, is:

Options:

(A) 1080

(B) 1134

(C) 1026

(D) 1188

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out into the first quadrant. Two lines, $L_1$ and $L_2$, stretch out from where you stand, creating a V-shape.

$L_1$ climbs steeply with a slope of $2$, while $L_2$ takes a more relaxed path with a slope of $\frac{1}{2}$.

Along these lines, points are scattered like stars in a constellation: $12$ points on $L_1$ and $9$ points on $L_2$. Including the origin itself, we have a total of $22$ points.

### The Strategy: Total Minus Invalid

When faced with a combinatorial problem like this, the most powerful tool in your arsenal is the **'Total minus Invalid'** strategy. Instead of trying to construct every possible triangle, we start by assuming every combination of $3$ points forms a triangle.

The total number of ways to select $3$ points from $22$ is given by the combination formula $^{22}C_{3}$.

$$^{22}C_{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 1540$$

This is our universe of possibilities. However, this universe contains 'imposters': sets of $3$ points that are collinear. These points lie on the same line and, therefore, cannot form a triangle.

### Identifying the Traps

Look closely at line $L_1$. It hosts the origin $O$ and the $12$ points $P_1, P_2, \dots, P_{12}$. That is $13$ points in total, all perfectly aligned.

Any selection of $3$ points from these $13$ will result in a flat line, not a triangle. The number of such invalid selections is:

$$^{13}C_{3} = \frac{13 	imes 12 	imes 11}{3 	imes 2 	imes 1} = 286$$

Similarly, line $L_2$ hosts the origin $O$ and the $9$ points $Q_1, Q_2, \dots, Q_9$, totaling $10$ collinear points. The number of invalid selections here is:

$$^{10}C_{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$$

### Final Calculation

Now, we simply subtract the invalid cases from our total. The number of valid triangles is $1540 - 286 - 120$.

Performing the arithmetic, we find:

$$1540 - 406 = 1134$$

By recognizing that the collinear points were the only obstacles, we transformed a daunting geometric problem into a clean, logical subtraction.

Remember this approach for your JEE exams: when the direct path is cluttered, look for the complement. You have successfully navigated the geometry of these lines and emerged with the correct count of **$1134$ triangles**.

Select Answer:

Visualized Solution

Visualizing the Geometry

Counting Total Points

The Triangle Formation Rule

Identifying Collinear Sets

Setting up the Equation

Calculating $^{22} C_{3}$

Calculating $^{13} C_{3}$

Calculating $^{10} C_{3}$

Final Subtraction

Conclusion & Key Takeaway

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Strategy

Identifying the Traps

Final Calculation

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