Maths Q7 – JEE-MAIN 2025

Home Maths Permutations and CombinationsQuestion #327

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}, . . ., P_{12}

are 12 points on line

L_{1}

and

Q_{1}, Q_{2}, . . ., 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}, . . ., P_{12}, Q_{1}, Q_{2}, . . ., Q_{9}

, is:

Clue 1 of 3: Identify all points

First, determine the total number of distinct points available for forming triangles.

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Step 1: Identify Total Points and Collinear Sets✦ Active

The total number of distinct points available is the origin O, 12 points on line $L_{1}$ ( $P_{1}, . . ., P_{12}$ ), and 9 points on line $L_{2}$ ( $Q_{1}, . . ., Q_{9}$ ). Thus, total points $N = 1 + 12 + 9 = 22$ .

The points on line $L_{1}$ are O and $P_{1}, . . ., P_{12}$ , making a set of $n_{1} = 1 + 12 = 13$ collinear points.

The points on line $L_{2}$ are O and $Q_{1}, . . ., Q_{9}$ , making a set of $n_{2} = 1 + 9 = 10$ collinear points.

Step 2: Calculate Total Combinations and Collinear Combinations○ Expand

The total number of ways to choose 3 points from 22 is given by $(\binom{22}{3})$ .

(\binom{22}{3}) = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 11 \times 7 \times 20 = 1540

The number of ways to choose 3 collinear points from the 13 points on $L_{1}$ is $(\binom{13}{3})$ .

(\binom{13}{3}) = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286

The number of ways to choose 3 collinear points from the 10 points on $L_{2}$ is $(\binom{10}{3})$ .

(\binom{10}{3}) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120

💡 Teacher's Secret Hint

Remember that the origin O is common to both lines and is included in the count for collinear points on each line.

Step 3: Calculate the Number of Triangles○ Expand

The total number of triangles is the total combinations of 3 points minus the combinations of 3 collinear points on $L_{1}$ and $L_{2}$ .

Number of triangles = (\binom{22}{3}) - (\binom{13}{3}) - (\binom{10}{3})

Number of triangles = 1540 - 286 - 120 = 1254 - 120 = 1134

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Maths Question 7 – JEE-MAIN 2025

Clue 1 of 3: Identify all points

🧠 Full Solution Path