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9.Straight Line
hard
Let the line $x+y=1$ meet the axes of $x$ and $y$ at $A$ and $B$, respectively. A right angled triangle $A M N$ is inscribed in the triangle $O A B$, 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 $AN : NB =\lambda: 1$, then the sum of all possible value$(s)$ of is $\lambda$ :
A$\frac{1}{2}$
B$\frac{13}{6}$
C$2$
D$\frac{5}{2}$
(JEE MAIN-2025)
Solution
imageArea of $\triangle AOB =\frac{1}{2}$
Area of $\triangle AMN =\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$
Equation of $A B$ is $x+y=1$
$OA=1, AM=\sec \left(45^{\circ}-\theta\right)$
$AN=\sec \left(45^{\circ}-\theta\right) \cos \theta$
$MN=\sec \left(45^{\circ}-\theta\right) \sin \theta$
$\operatorname{Ar}(\triangle AMN )=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9}$
$\Rightarrow \tan \theta=2, \frac{1}{2}$
$\tan \theta=2 \text { is rejected }$
$\frac{ AN }{ NB }=\frac{\lambda}{1}=\cot \theta=2$
Standard 11
Mathematics
