A variable line L passes through the point (3, 5) and intersects the positive coordinate axes at the

Q: Wait, why can't we just use calculus to minimize the area?

Calculus is a perfectly valid tool! You could express $b$ in terms of $a$ from the constraint, substitute it into the area formula, and find the derivative. However, the AM-GM inequality is often faster and more elegant for these types of problems, especially in competitive exams where time is precious.

Q: Why is the area of the triangle $\frac{1}{2}ab$?

The triangle is formed by the origin $(0, 0)$, the point $A(a, 0)$ on the x-axis, and the point $B(0, b)$ on the y-axis. Since the axes are perpendicular, this is a right-angled triangle. The base is the distance from the origin to $A$, which is $a$, and the height is the distance from the origin to $B$, which is $b$. Thus, $Area = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}ab$.

Q: What if the point was in a different quadrant, like $(3, -5)$?

The logic would change significantly. The problem specifies that the line intersects the **positive** coordinate axes, which implies $a > 0$ and $b > 0$. If the point were in a different quadrant, the intercepts might not be positive, and the AM-GM inequality (which requires positive numbers) would not be directly applicable in the same way. Always pay attention to the constraints given in the problem!

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 :

Options:

(A) 30

(B) 25

(C) 40

(D) 35

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a fixed point $P$ at $(3, 5)$. A line passes through this point, but it is not static; it is a variable line, meaning it can rotate around $P$.

However, it is constrained by a beautiful condition: it must intersect the positive x-axis and the positive y-axis. This creates a triangle $	riangle OAB$ with the origin $O(0, 0)$. Our goal is to find the minimum area this triangle can possibly have as the line rotates.

### The Intercept Form

To tackle this, we need to define our line mathematically. Let the line intersect the x-axis at $A(a, 0)$ and the y-axis at $B(0, b)$. Since the intercepts are on the positive axes, we know $a > 0$ and $b > 0$.

The most powerful tool for a line defined by its intercepts is the intercept form of the linear equation:

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

This equation is the heartbeat of our problem. It tells us exactly how the line behaves relative to the axes.

### The Constraint of the Fixed Point

We know the line must pass through the fixed point $P(3, 5)$. This means that the coordinates $(3, 5)$ must satisfy our intercept equation. By substituting $x = 3$ and $y = 5$ into the equation, we get:

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

This is our constraint. It links the two variables $a$ and $b$ together. No matter how the line rotates, this relationship must hold true.

### The Optimization Challenge

We want to minimize the area of the right-angled triangle $	riangle OAB$. The area is given by the classic formula:

$$Area = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2}ab$$

We have a sum constraint, $\frac{3}{a} + \frac{5}{b} = 1$, and we want to minimize the product $ab$. This is a classic setup for the **Arithmetic Mean-Geometric Mean (AM-GM) inequality**.

The AM-GM inequality states that for any positive real numbers $x_1$ and $x_2$, the arithmetic mean is greater than or equal to the geometric mean:

$$\frac{x_1 + x_2}{2} \ge \sqrt{x_1 x_2}$$

### Applying the AM-GM Magic

Let us apply this to our terms $\frac{3}{a}$ and $\frac{5}{b}$:

$$\frac{\frac{3}{a} + \frac{5}{b}}{2} \ge \sqrt{\frac{3}{a} \cdot \frac{5}{b}}$$

We know that $\frac{3}{a} + \frac{5}{b} = 1$. Substituting this into our inequality, we get:

$$\frac{1}{2} \ge \sqrt{\frac{15}{ab}}$$

To solve for $ab$, we square both sides. Since both sides are positive, the inequality sign remains unchanged:

$$\left(\frac{1}{2}ight)^2 \ge \frac{15}{ab}$$

$$\frac{1}{4} \ge \frac{15}{ab}$$

Cross-multiplying, we find:

$$ab \ge 15 	imes 4 = 60$$

### The Final Result

We have found that the product $ab$ must be at least 60. Since the area of our triangle is $\frac{1}{2}ab$, the minimum area is:

$$Area_{min} = \frac{1}{2} 	imes 60 = 30$$

And there we have it! The **minimum area of the triangle is 30**.

As a parting gift, remember this shortcut for your exams: for a line passing through $(h, k)$ in the first quadrant, the minimum area of the triangle formed with the axes is always $2hk$. Here, $2 	imes 3 	imes 5 = 30$. Keep this in your toolkit, and you will be ready for any similar challenge!

Animated Solution for Mathematics - Straight Lines: 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 $O A B$ , where $O$ is the origin, is :

Select Answer:

Visualized Solution

Visualizing the Setup

Intercept Form of the Line

Substituting the Fixed Point

Area of $△ O A B$

AM-GM Inequality

Applying AM-GM

Substituting the Sum

Squaring Both Sides

Finding the Minimum of $ab$

Calculating Minimum Area

Final Conclusion & Shortcut

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Intercept Form

The Constraint of the Fixed Point

The Optimization Challenge

Applying the AM-GM Magic

The Final Result

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