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 \frac49 of the area of the triangle OAB and AN : NB = \lambda : 1 , then the sum of all possible value(s) of \lambda is:

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q33 with Solution

View written solutionFree

Correct answer: B, $\FRAC{5}{2}$

Set up the figure

The line $x+y=1$ meets:

the $x$ -axis at $A=(1,0)$ ,
the $y$ -axis at $B=(0,1)$ .

Also, $O=(0,0)$ .

So triangle $OAB$ has area

[OAB]=\frac12.

We are told that triangle $AMN$ is right-angled, with:

$M$ on $OB$ ,
$N$ on $AB$ .

Since $OB$ is the segment on the line $x=0$ , let

M=(0,m), \qquad 0\le m\le 1.

Since $N$ lies on $AB$ , whose equation is $x+y=1$ , write

N=(t,1-t), \qquad 0\le t\le 1.

Use the area condition

Area of triangle $AMN$ is

[AMN]=\frac12\left|\det\begin{pmatrix}M-A & N-A\end{pmatrix}\right|.

Now,

M-A=(-1,m), \qquad N-A=(t-1,1-t).

Hence

\det\begin{pmatrix}-1 & t-1\\ m & 1-t\end{pmatrix} =(-1)(1-t)-m(t-1) =-(1-t)+m(1-t) =(1-t)(m-1).

[AMN]=\frac12(1-t)(1-m).

Given

[AMN]=\frac49[OAB]=\frac49\cdot\frac12=\frac29.

Thus

\frac12(1-t)(1-m)=\frac29 \implies (1-t)(1-m)=\frac49. \tag{1}

Use the right-angle condition

The right angle can occur at $M$ or at $N$ (it cannot be at $A$ in general unless specially forced, and we check the relevant possibilities).

Case 1: Right angle at $M$

Then

\overrightarrow{MA}\cdot\overrightarrow{MN}=0.

Now,

\overrightarrow{MA}=(1,-m), \qquad \overrightarrow{MN}=(t,1-t-m).

(1,-m)\cdot (t,1-t-m)=0

\implies t-m(1-t-m)=0

\implies t-m+mt+m^2=0.

A simpler geometric way: since $OB$ is vertical, for angle at $M$ to be $90^\circ$ , $MN$ must be horizontal. Hence $N$ must have the same $y$ -coordinate as $M$ :

1-t=m \implies t=1-m.

Substitute into (1):

(1-(1-m))(1-m)=m(1-m)=\frac49.

m-m^2=\frac49 \implies 9m^2-9m+4=0.

Its discriminant is

\Delta=81-144=-63<0,

so no real solution.

Hence right angle at $M$ is impossible.

Case 2: Right angle at $N$

Then

\overrightarrow{NA}\cdot\overrightarrow{NM}=0.

Since $A$ and $N$ lie on $AB$ whose slope is $-1$ , the line $AN$ has slope $-1$ . Therefore a perpendicular through $N$ must have slope $+1$ .

Thus line $NM$ must have slope $1$ .

But $M=(0,m)$ and $N=(t,1-t)$ , so slope of $NM$ is

\frac{m-(1-t)}{0-t}.

Set it equal to $1$ :

\frac{m-1+t}{-t}=1 \implies m-1+t=-t \implies m=1-2t.

1-m=2t.

Substitute into (1):

(1-t)(1-m)=(1-t)(2t)=\frac49.

Hence

2t(1-t)=\frac49

\implies 18t-18t^2=4

\implies 9t^2-9t+2=0.

Solve:

t=\frac{9\pm\sqrt{81-72}}{18}=\frac{9\pm 3}{18}.

t=\frac13 \quad \text{or} \quad t=\frac23.

Now $AN:NB=\lambda:1$ .

Since $N$ divides segment $AB$ with parameter $t$ , we have

N=(1-s)A+sB=(1-s,s).

Comparing with $N=(t,1-t)$ gives

s=1-t.

Thus

\frac{AN}{NB}=\frac{s}{1-s}=\frac{1-t}{t}.

\lambda=\frac{1-t}{t}.

For $t=\frac13$ ,

\lambda=\frac{2/3}{1/3}=2.

For $t=\frac23$ ,

\lambda=\frac{1/3}{2/3}=\frac12.

Hence all possible values are

\lambda=2,\ \frac12.

Their sum is

2+\frac12=\frac52.

Compare with options

The sum of all possible values of $\lambda$ is

\boxed{\frac52}.

So the correct option is B.

Comparison with stored answer

Stored correct answer is C (i.e. $2$ ), but our derived answer is B (i.e. $\frac52$ ).

Therefore, the stored answer appears to be incorrect.

More from Straight Lines and Pair of Straight Lines

View written solutionFree

Correct answer: B, $\FRAC{5}{2}$

Set up the figure

The line $x+y=1$ meets:

the $x$ -axis at $A=(1,0)$ ,
the $y$ -axis at $B=(0,1)$ .

Also, $O=(0,0)$ .

So triangle $OAB$ has area

[OAB]=\frac12.

We are told that triangle $AMN$ is right-angled, with:

$M$ on $OB$ ,
$N$ on $AB$ .

Since $OB$ is the segment on the line $x=0$ , let

M=(0,m), \qquad 0\le m\le 1.

Since $N$ lies on $AB$ , whose equation is $x+y=1$ , write

N=(t,1-t), \qquad 0\le t\le 1.

Use the area condition

Area of triangle $AMN$ is

[AMN]=\frac12\left|\det\begin{pmatrix}M-A & N-A\end{pmatrix}\right|.

Now,

M-A=(-1,m), \qquad N-A=(t-1,1-t).

Hence

\det\begin{pmatrix}-1 & t-1\\ m & 1-t\end{pmatrix} =(-1)(1-t)-m(t-1) =-(1-t)+m(1-t) =(1-t)(m-1).

[AMN]=\frac12(1-t)(1-m).

Given

[AMN]=\frac49[OAB]=\frac49\cdot\frac12=\frac29.

Thus

\frac12(1-t)(1-m)=\frac29 \implies (1-t)(1-m)=\frac49. \tag{1}

Use the right-angle condition

The right angle can occur at $M$ or at $N$ (it cannot be at $A$ in general unless specially forced, and we check the relevant possibilities).

Case 1: Right angle at $M$

Then

\overrightarrow{MA}\cdot\overrightarrow{MN}=0.

Now,

\overrightarrow{MA}=(1,-m), \qquad \overrightarrow{MN}=(t,1-t-m).

(1,-m)\cdot (t,1-t-m)=0

\implies t-m(1-t-m)=0

\implies t-m+mt+m^2=0.

A simpler geometric way: since $OB$ is vertical, for angle at $M$ to be $90^\circ$ , $MN$ must be horizontal. Hence $N$ must have the same $y$ -coordinate as $M$ :

1-t=m \implies t=1-m.

Substitute into (1):

(1-(1-m))(1-m)=m(1-m)=\frac49.

m-m^2=\frac49 \implies 9m^2-9m+4=0.

Its discriminant is

\Delta=81-144=-63<0,

so no real solution.

Hence right angle at $M$ is impossible.

Case 2: Right angle at $N$

Then

\overrightarrow{NA}\cdot\overrightarrow{NM}=0.

Since $A$ and $N$ lie on $AB$ whose slope is $-1$ , the line $AN$ has slope $-1$ . Therefore a perpendicular through $N$ must have slope $+1$ .

Thus line $NM$ must have slope $1$ .

But $M=(0,m)$ and $N=(t,1-t)$ , so slope of $NM$ is

\frac{m-(1-t)}{0-t}.

Set it equal to $1$ :

\frac{m-1+t}{-t}=1 \implies m-1+t=-t \implies m=1-2t.

1-m=2t.

Substitute into (1):

(1-t)(1-m)=(1-t)(2t)=\frac49.

Hence

2t(1-t)=\frac49

\implies 18t-18t^2=4

\implies 9t^2-9t+2=0.

Solve:

t=\frac{9\pm\sqrt{81-72}}{18}=\frac{9\pm 3}{18}.

t=\frac13 \quad \text{or} \quad t=\frac23.

Now $AN:NB=\lambda:1$ .

Since $N$ divides segment $AB$ with parameter $t$ , we have

N=(1-s)A+sB=(1-s,s).

Comparing with $N=(t,1-t)$ gives

s=1-t.

Thus

\frac{AN}{NB}=\frac{s}{1-s}=\frac{1-t}{t}.

\lambda=\frac{1-t}{t}.

For $t=\frac13$ ,

\lambda=\frac{2/3}{1/3}=2.

For $t=\frac23$ ,

\lambda=\frac{1/3}{2/3}=\frac12.

Hence all possible values are

\lambda=2,\ \frac12.

Their sum is

2+\frac12=\frac52.

Compare with options

The sum of all possible values of $\lambda$ is

\boxed{\frac52}.

So the correct option is B.

Comparison with stored answer

Stored correct answer is C (i.e. $2$ ), but our derived answer is B (i.e. $\frac52$ ).

Therefore, the stored answer appears to be incorrect.

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