Question

Through the point P(a, b) where ab> 0, the straight line = 1 is drawn so as to form with axes a triangle of area

S. If ab> 0 then least value of S is

• A. ab
• B. 2ab
• C. 3ab
• D. None

Answer 1

Answer: (B)

2ab

Answer 2

Area of DOAB = S = $\frac { 1 } { 2 }$ ab …….(i)

equation of AB is $\frac { \mathrm { x } } { \mathrm { a } } + \frac { \mathrm { y } } { \mathrm { b } }$ = 1

put (a, b)

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

Ž $\frac { \alpha } { a } + \frac { a \beta } { 2 S } = 1$ [using (i)]

Ž a²b – 2aS + 2aS = 0

\ aĪR

Ž D ³ 0

4S² – 8abS ³ 0

S ³ 2ab

Least value of S = 2ab