Key Concepts and Formulas

• Intercept Form of a Straight Line: If a line intersects the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$, its equation is given by: $\frac{x}{a} + \frac{y}{b} = 1$

• Area of Triangle OAB: For a line intersecting the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$, the area of the triangle OAB is: $\text{Area} = \frac{1}{2}ab$

• AM-GM Inequality: For non-negative real numbers $x_1, x_2, \dots, x_n$, the Arithmetic Mean is greater than or equal to the Geometric Mean: $\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$ Equality holds if and only if $x_1 = x_2 = \dots = x_n$.

Step-by-Step Solution

Step 1: Set up the equation of the line using the intercept form.

Let the line $L$ intersect the positive x-axis at point $A(a,0)$ and the positive y-axis at point $B(0,b)$. Since the line intersects the positive coordinate axes, we have $a > 0$ and $b > 0$. The equation of the line in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$

Step 2: Use the given point to establish a relationship between the intercepts.

The line $L$ passes through the point $(3,5)$. This means the coordinates $(3,5)$ satisfy the equation of the line. Substituting $x=3$ and $y=5$ into the line equation: $\frac{3}{a} + \frac{5}{b} = 1$ This equation relates $a$ and $b$, and we will use it as a constraint for optimization.

Step 3: Express the Area of triangle OAB in terms of the intercepts.

The area of triangle OAB, denoted by $A$, is given by: $A = \frac{1}{2}ab$ Our goal is to find the minimum value of this area, subject to the constraint $\frac{3}{a} + \frac{5}{b} = 1$.

Step 4: Minimize the Area using the AM-GM Inequality.

We have the constraint $\frac{3}{a} + \frac{5}{b} = 1$. We want to minimize the product $ab$. Since $a>0$ and $b>0$, the terms $\frac{3}{a}$ and $\frac{5}{b}$ are positive. Applying the AM-GM inequality to these two terms: $\frac{\frac{3}{a} + \frac{5}{b}}{2} \ge \sqrt{\frac{3}{a} \cdot \frac{5}{b}}$ Substitute the constraint $\frac{3}{a} + \frac{5}{b} = 1$ into the left side: $\frac{1}{2} \ge \sqrt{\frac{15}{ab}}$ Square both sides of the inequality: $\frac{1}{4} \ge \frac{15}{ab}$ Rearrange the inequality to find the minimum value of $ab$: $ab \ge 60$ The minimum value of $ab$ is 60. Therefore, the minimum area of triangle OAB is: $A_{min} = \frac{1}{2} (ab)_{min} = \frac{1}{2} (60) = 30$ The equality in the AM-GM inequality holds when $\frac{3}{a} = \frac{5}{b}$. Since their sum is 1, each term must be equal to $\frac{1}{2}$: $\frac{3}{a} = \frac{1}{2} \implies a = 6$ $\frac{5}{b} = \frac{1}{2} \implies b = 10$ Both $a=6$ and $b=10$ are positive, and satisfy the given conditions.

Step 5: State the alternative general result for minimum area.

For a line passing through a point $(x_0, y_0)$ and intersecting the positive coordinate axes at $(a,0)$ and $(0,b)$, the minimum area of the triangle OAB is $2x_0 y_0$. In this case, $(x_0, y_0) = (3,5)$. Minimum Area $= 2 \times 3 \times 5 = 30$.

Common Mistakes & Tips

• Domain Constraints: Remember the conditions $a>0$ and $b>0$.
• AM-GM Equality Condition: Check that the equality condition of the AM-GM inequality is achievable within the problem constraints.

Summary

The problem asked for the minimum area of the triangle formed by a line passing through $(3,5)$ and the positive coordinate axes. We used the intercept form of the line and applied the AM-GM inequality to find the minimum area, which is 30 square units.

The final answer is $\boxed{30}$, which corresponds to option (C).