JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Straight Lines

Let the line

x + y = 1

meet the axes of

x

and

y

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 = λ : 1

, then the sum of all possible value(s) of is

λ

(A) $\frac{1}{2}$
(B) $\frac{13}{6}$
(C) $\frac{5}{2}$
(D)2✓ Correct

Answer: (D)

Solution

A = (1, 0), B = (0, 1)

, area of

△ O A B = \frac{1}{2}

, so area of

△ A M N = \frac{4}{9} \cdot \frac{1}{2} = \frac{2}{9}

.

With the right angle at

N

(on

A B

), let

θ

describe the direction of

A N

relative to

A B

, so that

∠ N A B = θ

and

∠ N A M = 45° - θ

. With

O A = 1

and

M

O B

A M = sec (45° - θ), A N = sec (45° - θ) cos θ, M N = sec (45° - θ) sin θ

Area (△ A M N) = \frac{1}{2} sec^{2} (45° - θ) sin θ cos θ = \frac{2}{9}

Solving this gives

tan θ = 2

tan θ = \frac{1}{2}

; the value

tan θ = 2

places

N

outside segment

A B

and is rejected, leaving

tan θ = \frac{1}{2}

λ = \frac{A N}{N B} = cot θ = 2

Since this is the only admissible value, the sum of all possible values of

λ

2

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Straight Lines

Solution

More JEE Main questions on Straight Lines