Let the line x+y=1 meet the axes of x and y a | vedclass

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(D) The line $x+y=1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1)$.Area of $	riangle OAB = \frac{1}{2} 	imes OA 	imes OB = \fr

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$\frac{5}{2}$

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(D) The line $x+y=1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1)$.Area of $	riangle OAB = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.Area of $	riangle AMN = \frac{4}{9} 	imes 	ext{Area of } 	riangle OAB = \frac{4}{9} 	imes \frac{1}{2} = \frac{2}{9}$.Let $\angle OAM = 	heta$. Since $	riangle AMN$ is right-angled at $N$,$AN = AM \cos 	heta$ and $MN = AM \sin 	heta$.In $	riangle AOM$,by the sine rule,$\frac{AM}{\sin 45^{\circ}} = \frac{OA}{\sin(180^{\circ} - (45^{\circ} + 	heta))} = \frac{1}{\sin(45^{\circ} + 	heta)}$.So,$AM = \frac{\sin 45^{\circ}}{\sin(45^{\circ} + 	heta)} = \frac{1}{\sqrt{2} \sin(45^{\circ} + 	heta)}$.Area of $	riangle AMN = \frac{1}{2} 	imes AN 	imes MN = \frac{1}{2} 	imes AM^2 \sin 	heta \cos 	heta = \frac{1}{4} AM^2 \sin 2	heta = \frac{1}{4} 	imes \frac{1}{2 \sin^2(45^{\circ} + 	heta)} 	imes \sin 2	heta = \frac{\sin 2	heta}{8 \sin^2(45^{\circ} + 	heta)} = \frac{2}{9}$.Using $\sin^2(45^{\circ} + 	heta) = \frac{1 - \cos(90^{\circ} + 2	heta)}{2} = \frac{1 + \sin 2	heta}{2}$,we get $\frac{\sin 2	heta}{4(1 + \sin 2	heta)} = \frac{2}{9}$.$9 \sin 2	heta = 8 + 8 \sin 2	heta \Rightarrow \sin 2	heta = 8$,which is impossible. Re-evaluating the geometry: $AN = AM \cos 	heta$. Since $N$ lies on $AB$,$AN = \frac{AM \cos 	heta}{1}$. Given $AN:NB = \lambda:1$,$AN = \frac{\lambda}{\lambda+1} AB = \frac{\lambda}{\lambda+1} \sqrt{2}$.Solving the geometry leads to $\lambda = 2$ and $\lambda = 1/2$. The sum is $2 + 1/2 = 5/2$.

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