Question

Through the point P(α, β), where αβ> 0, the straight line $\frac { x } { a } + \frac { y } { b } = 1$ is drawn so as to form with coordinate axes a triangle of area s. If ab> 0, then least value of s is

• A. 2αβ
• B. 1/2 αβ
• C. αβ
• D. None of these

Answer 1

Answer: (A)

2αβ

Answer 2

Given P≡ (α, β).

Given line is $\frac { x } { a } + \frac { y } { b } = 1$ .. . . . . (1)

If line (1) cuts x and y axes at A and B respectively, then

A≡ (a, 0) and B ≡ (0, b). Also the area of ∆OAB = s

i.e. (1/2)ab = s ⇒ ab = 2s.

Since line (1) passes through P(α, β),

$\frac { \alpha } { a } + \frac { \beta } { b } = 1$⇒ $\frac { \alpha } { a } + \frac { a \beta } { 2 s } = 1$ ⇒ a² β – 2as + 2αs = 0.

Since a is real, 4s² – 8αβs ≥ 0 ⇒ s ≥ 2αβ .

Hence the least value of s = 2αβ.

Hence (1) is the correct answer.

Alternative solution:

Since (α, β) lies on the given line, $\frac { \alpha } { a } + \frac { \beta } { b } = 1$

⇒ ab = aβ + bα ≥ 2 ⇒ 2s = ab ≥ 4αβ

⇒ least value of s = 2αβ.

Hence (1) is the correct answer