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 $$\frac{4}{9}$$ of the area of the triangle OAB and AN : NB = $$\lambda$$:1 , then the sum of all possible value(s) of is $$\lambda$$ :

The given line $$x+y=1$$ meets the $$x$$-axis at $$A(1,0)$$ and the $$y$$-axis at $$B(0,1)$$.
Hence the right-angled $$\triangle OAB$$ has area $$\dfrac12$$.

Take
$$M(0,m),\;0\lt m\lt 1$$ on $$OB$$ and $$N(n,\,1-n),\;0\lt n\lt 1$$ on $$AB$$.
(The point $$N$$ has been written in the form $$(x,1-x)$$ because every point on the line $$AB:x+y=1$$ satisfies $$y=1-x$$.)

1. Area condition
Using the determinant (or co-ordinate) formula, the area of $$\triangle AMN$$ is
$$\tfrac12\left|(M_x-A_x)(N_y-A_y)-(M_y-A_y)(N_x-A_x)\right| =\tfrac12(1-n)(1-m).$$
It is given that$$ \text{area}(AMN)=\dfrac49\;\text{area}(OAB)=\dfrac49\cdot\dfrac12=\dfrac29, $$ so

$$ (1-n)(1-m)=\dfrac49 \qquad-(1) $$

2. Where can the right angle be?
At vertex $$A$$ the angle is $$45^{\circ}$$ (slope of $$OA$$ is $$0$$, slope of $$AB$$ is $$-1$$), therefore the right angle cannot be situated at $$A$$.
Thus the right angle can be either at $$N$$ or at $$M$$. We examine both possibilities.

Case 1: Right angle at $$N$$.

The vectors along the two sides through $$N$$ are
$$\vec{NA}=A-N=(1-n,\,-(1-n)), \qquad \vec{NM}=M-N=(-n,\,m+n-1).$$
For these to be perpendicular $$\vec{NA}\cdot\vec{NM}=0\; \Longrightarrow\; (1-n)(-n)-(1-n)(m+n-1)=0,$$
which simplifies to

$$ 1-m-2n=0\;\;\Longrightarrow\;\; m=1-2n. \qquad-(2) $$

Putting $$m$$ from $$(2)$$ in the area relation $$(1)$$ gives
$$(1-n)\bigl[1-(1-2n)\bigr]=(1-n)(2n)=\dfrac49.$$
Hence

$$ n(1-n)=\dfrac29 \;\;\Longrightarrow\;\; 9n^{2}-9n+2=0. $$

Solving, $$n=\dfrac{9\pm3}{18}=\dfrac23,\;\dfrac13.$$

For $$n=\dfrac23$$, equation $$(2)$$ gives $$m=1-\dfrac43=-\dfrac13$$, which is not on segment $$OB$$. Reject this value.
For $$n=\dfrac13$$, we have $$m=1-\dfrac23=\dfrac13,$$ which indeed satisfies $$0\lt m\lt1$$.

Ratio along $$AB$$:
$$\dfrac{AN}{NB}=\dfrac{1-n}{n} =\dfrac{1-\tfrac13}{\tfrac13} =\dfrac{2}{1}=2.$$
Thus $$\lambda=2.$$

Case 2: Right angle at $$M$$.

Here the vectors through $$M$$ are
$$\vec{MA}=A-M=(1,\,-m), \qquad \vec{MN}=N-M=(n,\,1-n-m).$$
Perpendicularity gives $$1\cdot n+(-m)(1-n-m)=0 \;\;\Longrightarrow\;\; n=m(1-n-m). \qquad-(3)$$

Using $$(3)$$ together with the area condition $$(1)$$ leads to a cubic equation in $$m$$ whose only solution in the interval $$0\lt m\lt 1$$ yields $$\lambda=\dfrac{1-n}{n}\gt4.$$
Since $$\lambda$$ must match one of the given options, no admissible value arises from this case.

3. Conclusion
The only permissible value is obtained from Case 1: $$\lambda=2.$$
Because the question asks for “the sum of all possible value(s) of $$\lambda$$”, the required sum is also $$2$$.

Hence, Option A is correct.