JEE Main 2025 — Straight Lines Question with Solution

$\begin{aligned} & \text { Area of } \triangle \mathrm{AOB}=\frac{1}{2} \\ & \text { Area of } \triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9} \end{aligned}$ Equation of AB is $\mathrm{x}+\mathrm{y}=1$ $\begin{aligned} & \mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right) \\ & \mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta \\ & \mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta \end{aligned}$ $\begin{array}{rl}& \mathrm{Ar}(△\mathrm{AMN})=\frac{1}{2}\times {\sec }^2\left(45^{\circ }-\theta \right)\sin \theta \cdot \cos \theta =\frac{2}{9}\\ & \Rightarrow \tan \theta =2,\frac{1}{2}\\ & \tan \theta =2\text{ is rejected }\\ & \frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda }{1}=\cot \theta =2\end{array}$