Question

A variable line $L$ passes through the point $(3, 5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is:

• A. 30
• B. 25
• C. 40
• D. 35

Answer 1

Answer: (A)

30

Answer 2

Step 1: Equation of the line L The line L passes through the point $(3, 5)$ and intersects the axes. Let the equation of the line L be:

$\frac{x}{a} + \frac{y}{b} = 1,$

where $a$ and $b$ are the intercepts on the $x$-axis and $y$-axis, respectively.

Since the line passes through $(3, 5)$, substitute $x = 3$ and $y = 5$:

$\frac{3}{a} + \frac{5}{b} = 1.$ (1)

Step 2: Area of triangle $OAB$ The area of triangle $OAB$ is given by:

$\text{Area} = \frac{1}{2} \times a \times b.$

From equation (1), express $b$ in terms of $a$:

$\frac{5}{b} = 1 - \frac{3}{a}$. $b = \frac{5a}{a - 3}.$ (2)

Substitute $b = \frac{5a}{a - 3}$ into the area formula:

$\text{Area} = \frac{1}{2} \times a \times \frac{5a}{a - 3}. \\\\\text{Area} = \frac{5a^2}{2(a - 3)}$ (3)

Step 3: Minimize the area Let $f(a) = \frac{5a^2}{2(a - 3)}$. To find the minimum area, calculate $\frac{df}{da}$ and set it equal to zero:

$f(a) = \frac{5a^2}{2(a - 3)}.$

Using the quotient rule:

$\frac{df}{da} = \frac{(2(a - 3)(10a)) - (5a^2(2))}{4(a - 3)^2}.$ $\frac{df}{da} = \frac{20a(a - 3) - 10a^2}{4(a - 3)^2}.$ $\frac{df}{da} = \frac{20a^2 - 60a - 10a^2}{4(a - 3)^2}.$ $\frac{df}{da} = \frac{10a(a - 6)}{4(a - 3)^2}.$

Set $\frac{df}{da} = 0$:

$10a(a - 6) = 0.$

Since $a = 0$ is not valid (intercept cannot be zero), $a = 6$.

Step 4: Calculate $b$ and the minimum area Substitute $a = 6$ into equation (2) to find $b$:

$b = \frac{5(6)}{6 - 3} = \frac{30}{3} = 10.$

The minimum area is:

$\text{Area} = \frac{1}{2} \times 6 \times 10 = 30.$

Final Answer: Option (1).