Let the line x + y = 1$x + y = 1$ meet the axes of x$x$ and y$y$ 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 frac49$\frac{4}{9}$ of the area of the triangle OAB and AN: NB = \lambda : 1, then the sum of all possible value(s) of lambda$\lambda$ is :

### Related Formula Area of a right-angled triangle: textArea = frac12 times textbase times textheight$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ ### Core Logic The line equation is x + y = 1$x + y = 1$, giving intercept coordinates A(1, 0)$A(1, 0)$ and B(0, 1)$B(0, 1)$.

Area of Delta OAB = frac12 times 1 times 1 = frac12$\Delta OAB = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$. Given area condition: textArea of Delta AMN = frac49 times frac12 = frac29$$\text{Area of } \Delta AMN = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$$ ### Step 1: Set up Trigonometric Tracing Let angle MAO = 45^circ - theta$\angle MAO = 45^\circ - \theta$. Since Delta OAB$\Delta OAB$ is isosceles right-angled at O$O$, angle OAB = 45^circ$\angle OAB = 45^\circ$. This establishes: OA = 1, quad AM = sec(45^circ - theta)$$OA = 1, \quad AM = \sec(45^\circ - \theta)$$ AN = sec(45^circ - theta)costheta$$AN = \sec(45^\circ - \theta)\cos\theta$$ MN = sec(45^circ - theta)sintheta$$MN = \sec(45^\circ - \theta)\sin\theta$$ ### Step 2: Solve for Angles and Ratios textArea(Delta AMN) = frac12 times sec^2(45^circ - theta)sinthetacostheta = frac29$$\text{Area}(\Delta AMN) = \frac{1}{2} \times \sec^2(45^\circ - \theta)\sin\theta\cos\theta = \frac{2}{9}$$ Solving the trigonometric ratio yields: tantheta = 2 quad textor quad frac12$$\tan\theta = 2 \quad \text{or} \quad \frac{1}{2}$$ Rejecting tantheta = 2$\tan\theta = 2$ based on physical boundaries within the triangle limits: fracANNB = fraclambda1 = cottheta = 2$$\frac{AN}{NB} = \frac{\lambda}{1} = \cot\theta = 2$$ Thus, the valid evaluation matches the option sequence value of 2. ### Pattern Recognition When dealing with inscribed right triangles inside symmetric linear bounds, parameterizing coordinates with angles matching the axis slope simplifies configuration variables dramatically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines