Let (h, k) be a fixed point, where h 0, k 0. A straight line passing through this point cuts the

Q: Why can't we just use the slope-intercept form $y = mx + c$?

You absolutely could, but it would make your life much harder! The slope-intercept form introduces $m$ and $c$ as variables, which are not directly related to the triangle's base and height. The intercept form, $\frac{x}{a} + \frac{y}{b} = 1$, is tailor-made for problems involving coordinate axes. It directly links the intercepts $a$ and $b$ to the point $(h, k)$, allowing you to solve the problem with much cleaner algebra.

Q: Is there a faster way to solve this than using calculus?

Yes, there is! The AM-GM inequality (Arithmetic Mean-Geometric Mean) is a powerful shortcut here. Since $\frac{h}{a} + \frac{k}{b} = 1$, and both terms are positive, we know that $\frac{\frac{h}{a} + \frac{k}{b}}{2} \ge \sqrt{\frac{hk}{ab}}$. Squaring both sides gives $\frac{1}{4} \ge \frac{hk}{ab}$, which simplifies to $ab \ge 4hk$. Since the area is $\frac{1}{2}ab$, the minimum area is $\frac{1}{2}(4hk) = 2hk$. Calculus is great, but recognizing these patterns is what makes you a JEE topper!

Q: Why do we assume $a > h$?

Great observation! If $a \le h$, the line would not be able to pass through $(h, k)$ while still cutting the positive x-axis at $a$ and the positive y-axis at $b$. If $a = h$, the line would be vertical, and if $a < h$, the line would intersect the y-axis at a negative value or not at all in the first quadrant. The math naturally enforces this constraint, but it's important to visualize why it's physically necessary.

Question

Let $(h, k)$ be a fixed point, where $h > 0, k > 0$. A straight line passing through this point cuts the positive direction of the coordinate axes at the points $P$ and $Q$. Find the minimum area of the triangle $OPQ, O$ being the origin.

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: A Journey to the Minimum

Imagine you are standing on a coordinate plane. You have a fixed anchor point at $(h, k)$ in the first quadrant.

You are holding a straight rod, and you must place it such that it passes through your anchor point and rests against the $x$ and $y$ axes, forming a triangle with the origin. As you rotate the rod, the triangle changes shape.

Your mission is to find the exact orientation of that rod that minimizes the area of the triangle $OPQ$.

### Phase 1: The Intercept Form

To begin, we need a mathematical language to describe our rod. Since the rod cuts the $x$-axis at $P(a, 0)$ and the $y$-axis at $Q(0, b)$, the most elegant way to describe this line is the intercept form:

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

This equation is beautiful because it tells us exactly where the line hits the axes. But our rod is not free; it is pinned at $(h, k)$.

This means the coordinates $(h, k)$ must satisfy our equation. Substituting these into our line equation, we get the constraint that governs our entire problem:

$$\frac{h}{a} + \frac{k}{b} = 1$$

This is the heartbeat of the problem. It links the intercepts $a$ and $b$ to our fixed constants $h$ and $k$.

### Phase 2: The Area Function

The area of our right-angled triangle $OPQ$ is $A = \frac{1}{2}ab$. We have a problem, though: $A$ depends on two variables, $a$ and $b$.

To minimize it, we need to express $A$ in terms of just one variable. Let's go back to our constraint: $\frac{k}{b} = 1 - \frac{h}{a}$.

Simplifying the right side, we get $\frac{k}{b} = \frac{a - h}{a}$. Flipping this, we find the expression for $b$:

$$b = \frac{ak}{a - h}$$

Now, substitute this into our area formula:

$$A(a) = \frac{1}{2} a \left( \frac{ak}{a - h} \right) = \frac{k}{2} \left( \frac{a^2}{a - h} \right)$$

We have successfully reduced the problem to a single-variable function, $A(a)$.

### Phase 3: The Calculus of Optimization

Now, we enter the realm of calculus. To find the minimum, we need to find where the rate of change of the area is zero.

We differentiate $A(a)$ with respect to $a$ using the quotient rule:

$$\frac{dA}{da} = \frac{k}{2} \left[ \frac{2a(a - h) - a^2(1)}{(a - h)^2} \right]$$

Let's simplify that numerator. Expanding the terms, we get $2a^2 - 2ah - a^2$, which simplifies beautifully to $a^2 - 2ah$.

So, our derivative is:

$$\frac{dA}{da} = \frac{k}{2} \left[ \frac{a^2 - 2ah}{(a - h)^2} \right]$$

For the area to be at a minimum, we set $\frac{dA}{da} = 0$. This implies $a^2 - 2ah = 0$.

Factoring this, we get $a(a - 2h) = 0$. Since $a$ must be positive, we discard $a=0$ and find our critical point: $a = 2h$.

### Phase 4: The Elegant Conclusion

With $a = 2h$, we can find the corresponding $y$-intercept $b$:

$$b = \frac{(2h)k}{2h - h} = \frac{2hk}{h} = 2k$$

Finally, we calculate the minimum area:

$$A_{min} = \frac{1}{2} (2h)(2k) = 2hk$$

Look at that result: **$2hk$**. It is so simple and symmetric.

There is a deeper geometric truth here: the minimum area occurs exactly when the fixed point $(h, k)$ is the midpoint of the segment $PQ$.

The midpoint of $P(2h, 0)$ and $Q(0, 2k)$ is indeed $(\frac{2h+0}{2}, \frac{0+2k}{2}) = (h, k)$.

Animated Solution for Mathematics - Straight Lines: Let $(h, k)$ be a fixed point, where $h > 0, k > 0$ . A straight line passing through this point cuts the positive direction of the coordinate axes at the points $P$ and $Q$ . Find the minimum area of the triangle $O P Q, O$ being the origin.

Visualized Solution

Setting up the Coordinate System

Drawing the Intersecting Line

Equation of the Line

Applying the Point Constraint

Expressing $b$ in terms of $a$

Area of Triangle $O P Q$

Area as a Function of $a$

Differentiating the Area Function

Simplifying the Derivative

Finding the Critical Point

Finding the Corresponding y-intercept

Calculating the Minimum Area

Geometric Interpretation

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

Phase 4

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