Line \(L_{1}\) of slope 2 and line \(L_{2}\) of slope \(\frac{1}{2}\) intersect at the origin O . In the first quadrant, \(\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots . \mathrm{P}_{12}\) are 12 points on line \(\mathrm{L}_{1}\) and \(\mathrm{Q}_{1}, \mathrm{Q}_{2}, \ldots \ldots \mathrm{Q}_{9}\) are 9 points on line \(\mathrm{L}_{2}\). Then the total number of triangles, that can be formed having vertices at three of the 22 points \(\mathrm{O}, \mathrm{P}_{1}, \mathrm{P}_{2}, \ldots \mathrm{P}_{12}\), \(\mathrm{Q}_{1}, \mathrm{Q}_{2}, \ldots . \mathrm{Q}_{9}\), is:

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Permutations and Combinations

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\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:

Answer: (B)

The

22

points comprise

13

L_{1}

(including

O

) and

10

L_{2}

(including

O

), with

O

common to both. Three points form a triangle unless they are collinear, i.e. all on

L_{1}

or all on

L_{2}

. Thus the number of triangles is

(3 22) - (3 13) - (3 10) = 1540 - 286 - 120

, which equals

1134

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option