JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 2)
Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin O . In the first quadrant, $\mathrm{P}_1$, $P_2, \ldots, P_{12}$ are 12 points on line $L_1$ and $Q_1, Q_2, \ldots, Q_9$ 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 $\mathrm{O}, \mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{12}$, $\mathrm{Q}_1, \mathrm{Q}_2, \ldots, \mathrm{Q}_9$, is:
1026
1188
1134
1080
Explanation
Total triangles
_3rd_April_Evening_Shift_en_2_1.png)
(2 points as $y=\frac{x}{2}, 1$ point on $y=\frac{x}{2}$)
$+2\left(\right.$ points and $y=\frac{x}{2}, 1$ point on $y=2 x$)
+(1 point on $y=2 x, 1)$ point on $y=\frac{x}{2}$ and origin)
$$={ }^9 C_2,{ }^{12} C_1+{ }^9 C_1{ }^{12} C_2+{ }^9 C_1 \cdot{ }^{12} C_1 \cdot{ }^1 C_1$$
$$=1134$$
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