Q52 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
A straight line L passes through the point of intersection of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. If L intersects the positive x-axis at $A(a, 0)$ and the positive y-axis at $B(0, b)$, then the minimum area of the triangle $OAB$ (where $O$ is the origin) is ....
- A$6$ square units.
- Correct answerB$12$ square units.
- C$24$ square units.
- D$48$ square units.
Solution
Solving the equations $x - y + 1 = 0$ and $2x + y - 7 = 0$, we get the point of intersection as $(2, 3)$.
Let the equation of the line $L$ be $\dfrac{x}{a} + \dfrac{y}{b} = 1$.
Since it passes through $(2, 3)$, we have:
$\dfrac{2}{a} + \dfrac{3}{b} = 1$
Since $L$ intersects the positive x-axis and positive y-axis, $a > 0$ and $b > 0$. Applying the AM-GM inequality on $\dfrac{2}{a}$ and $\dfrac{3}{b}$:
$\dfrac{\dfrac{2}{a} + \dfrac{3}{b}}{2} \ge \sqrt{\dfrac{2}{a} \times \dfrac{3}{b}}$
$\dfrac{1}{2} \ge \sqrt{\dfrac{6}{ab}}$
Squaring both sides:
$\dfrac{1}{4} \ge \dfrac{6}{ab} \Rightarrow ab \ge 24$
The area of the triangle $OAB$ is given by $\Delta = \dfrac{1}{2}ab$.
$\Delta \ge \dfrac{1}{2} \times 24 = 12$
The minimum area of the triangle is $12$ square units.
Let the equation of the line $L$ be $\dfrac{x}{a} + \dfrac{y}{b} = 1$.
Since it passes through $(2, 3)$, we have:
$\dfrac{2}{a} + \dfrac{3}{b} = 1$
Since $L$ intersects the positive x-axis and positive y-axis, $a > 0$ and $b > 0$. Applying the AM-GM inequality on $\dfrac{2}{a}$ and $\dfrac{3}{b}$:
$\dfrac{\dfrac{2}{a} + \dfrac{3}{b}}{2} \ge \sqrt{\dfrac{2}{a} \times \dfrac{3}{b}}$
$\dfrac{1}{2} \ge \sqrt{\dfrac{6}{ab}}$
Squaring both sides:
$\dfrac{1}{4} \ge \dfrac{6}{ab} \Rightarrow ab \ge 24$
The area of the triangle $OAB$ is given by $\Delta = \dfrac{1}{2}ab$.
$\Delta \ge \dfrac{1}{2} \times 24 = 12$
The minimum area of the triangle is $12$ square units.
Hence, the correct answer is option: 'B'.