Mathematics · Straight lines

JEE Main 2025 — 29 January, Evening Shift — Question 49

Let the line $x+y=1$ meet the axes of $x$ and $y$ at $A$ and B , respectively. A right angled triangle AMN is inscribed in the triangle OAB ,

where O is the origin and the points M and N lie on the lines OB and $A B$ , respectively. If the area of the triangle AMN is $\frac{4}{9}$ of the area of the triangle OAB and $\mathrm{AN}: \mathrm{NB}=\lambda: 1$ , then the sum of all possible value(s) of is $\lambda$

Option A:
$\frac{1}{2}$
Option B:
$\frac{13}{6}$
Option C:
$\frac{5}{2}$
Option D:
2
Correct

Answer: D

Step-by-step solution

Area of $\triangle \mathrm{AOB}=\frac{1}{2}$

Area of $\triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$

Equation of AB is $\mathrm{x}+\mathrm{y}=1$

$\mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right)$

$\mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta$

$\mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta$

$\operatorname{Ar}(\triangle \mathrm{AMN})=\frac{1}{2} \times \sec ^{2}\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9}$

$\Rightarrow \tan \theta=2, \frac{1}{2}$

$\tan \theta=2$ is rejected

$\frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda}{1}=\cot \theta=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Family of lines, optimisation.