Question

A straight line is drawn through the point P(3, 4) meeting the positive direction of co­ordinate axes at the points A and B. If '0' is the origin then minimum area of triangle OAB is equal to

• A. 12 sq. units
• B. 6 sq. units
• C. 24 sq. units
• D. 48 sq. units

Answer 1

Answer: (C)

24 sq. units

Answer 2

Let the equation of drawn line be$\frac { x } { a } + \frac { y } { b } = 1$

where a > 3, b > 4, as the line passes through (3,4) and meets the positive direction of co­ordinate axis. We have$\frac { 3 } { a } + \frac { 4 } { b } = 1$

⇒ b = $\frac { 4 a } { ( a - 3 ) }$. Now area of triangle OAB.

$\Delta = \frac { 1 } { 2 } a b = \frac { 2 a ^ { 2 } } { ( a - 3 ) } \cdot \frac { d \Delta } { d a } = \frac { 2 a ( a - 6 ) } { ( a - 3 ) ^ { 2 } }$

Clearly a = 6 is the point of minima for ∆.

Thus = 24 sq. units.