Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
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 $AB$, respectively. If the area of the triangle $AMN$ is $\dfrac{4}{9}$ of the area of the triangle $OAB$ and $AN:NB = \lambda:1$, then the sum of all possible values of $\lambda$ is:
$2$
$\dfrac{5}{2}$
$\dfrac{1}{2}$
$\dfrac{13}{6}$
Step-by-Step Solution
Key Concept: Express $\text{Area}(\triangle AMN)$ in terms of the angle $\theta$ that $AN$ makes with $AB$; set equal to $\tfrac{4}{9}\cdot\tfrac{1}{2}=\tfrac{2}{9}$; solve for $\theta$ and use $\lambda = \cot\theta$.
Area of $\triangle OAB = \tfrac{1}{2}$. Area of $\triangle AMN = \tfrac{4}{9}\times\tfrac{1}{2} = \tfrac{2}{9}$.
With $OA=1$, $AM = \sec(45^\circ-\theta)$, $AN = \sec(45^\circ-\theta)\cos\theta$, $MN = \sec(45^\circ-\theta)\sin\theta$:
$\text{Area}(\triangle AMN) = \tfrac{1}{2}\sec^2(45^\circ-\theta)\sin\theta\cos\theta = \tfrac{2}{9}$.
$\Rightarrow \tan\theta = 2$ (rejected, puts $N$ outside $AB$) or $\tan\theta = \tfrac{1}{2}$.
$\lambda = AN/NB = \cot\theta = 2$. Sum of valid values $= 2$.
Correct Answer: 1