Through the point P(α , β), where α β > 0, the straight line $\large \frac{x}{a} + \frac{y}{b} = 1$ is

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

(A) 2α β

(B) (1/2 ) α β

(C) α β

(D) none of these

Ans: (A)

Sol: Given P ≡ (α , β)

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

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

A ≡ (a, 0) and B ≡ (0, b).

Also the area of ΔOAB = s

(1/2)ab = s ⇒  ab = 2s

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

$\large \frac{\alpha}{a} + \frac{\beta}{b} = 1$

$\large \frac{\alpha}{a} + \frac{\alpha \beta}{2 s} = 1$

⇒ a² β – 2a s + 2α s = 0

Since a is real, 4s² – 8 α β s  ≥ 0

⇒  s ≥ 2 α β  .

Hence the least value of  s = 2 α β