JEE Mathematics 2025 Question with Solution

Given: There are $12$ points on line $L_1$, $9$ points on line $L_2$, and both lines pass through the origin $O$.

Find: The total number of triangles formed by choosing any $3$ of the $22$ given points.

First count all possible selections of $3$ points from $22$ points:

Now exclude all collinear selections.

On line $L_1$, the $12$ points $P_1, P_2, \ldots, P_{12}$ are collinear, so:

On line $L_2$, the $9$ points $Q_1, Q_2, \ldots, Q_9$ are collinear, so:

Also, the origin $O$ lies on both lines.

So the selections $\left(O, P_i, P_j\right)$ are collinear on $L_1$:

And the selections $\left(O, Q_k, Q_l\right)$ are collinear on $L_2$:

Hence total collinear selections are:

Therefore, the number of triangles is:

Therefore, the total number of triangles formed is $1134$. The correct option is B.

Given: Points lie on two lines through the origin: $12$ points on $L_1$, $9$ points on $L_2$, and the origin $O$.

Find: The number of non-collinear triples.

Count directly by valid triangle types.

Case 1: Choose $2$ points from $L_1$ and $1$ point from $L_2$:

Case 2: Choose $1$ point from $L_1$ and $2$ points from $L_2$:

Case 3: Choose the origin $O$, $1$ point from $L_1$, and $1$ point from $L_2$:

Adding all valid cases:

Therefore, the total number of triangles is $1134$, so the correct option is B.