Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angl

Q: Wait, why does the distance $d$ remain the same? Doesn't rotating the axes change the coordinates of the line?

That is a brilliant question! It is easy to get confused here. Remember, the line $L$ is a fixed physical entity in the plane. The origin $(0,0)$ is also fixed. The distance $d$ is the length of the perpendicular segment dropped from the origin to the line. Since neither the origin nor the line has moved, the length of that segment cannot possibly change. The rotation of axes only changes the *description* of the line (its intercepts), not its *position*.

Q: Can't we just use rotation matrices to transform the coordinates and solve this? It seems more rigorous.

You absolutely *could* use rotation matrices! You would transform $(x, y)$ to $(x', y')$ using $x = x'\cos\theta - y'\sin\theta$ and $y = x'\sin\theta + y'\cos\theta$. However, that path is a labyrinth of trigonometry. In JEE Advanced, the goal is not just to solve, but to solve *elegantly*. Recognizing the invariance of the perpendicular distance saves you pages of calculation and minimizes the chance of a sign error.

Q: Is the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ always valid? What if the line passes through the origin?

You are thinking like a mathematician! You are correct: the intercept form is undefined if the line passes through the origin (because $a$ or $b$ would be zero). However, in this problem, the line has intercepts $a$ and $b$, which implies it does *not* pass through the origin. If it did, the distance $d$ would be zero, and the equation would be trivial. Always check the domain of your formulas, but here, the setup guarantees the intercepts exist.

Question

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$, then

Options:

(A) $a^2 + b^2 = p^2 + q^2$

(B) $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$

(C) $a^2 + p^2 = b^2 + q^2$

(D) $\frac{1}{a^2} + \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{q^2}$

Sigma Percentile · Accepted Answer

### The Geometry of Invariance: A Masterclass in Perspective

Imagine you are standing in a room, holding a rigid, straight metal rod. You place this rod against the corner of the room, such that it rests against the two walls.

In our coordinate geometry world, the corner is the origin $(0,0)$, and the two walls are the $x$-axis and the $y$-axis. The points where the rod touches the walls are the intercepts, $a$ and $b$. This is the physical reality of our problem.

Now, suppose you decide to rotate your entire perspective. You don't move the rod—it stays exactly where it is, suspended in the air. You simply rotate your head, or perhaps you rotate the entire room around the origin.

Suddenly, the walls are at a different angle. The rod now touches the 'new' walls at points $p$ and $q$. The question asks us to find the relationship between these intercepts. This is not just an algebra problem; it is a meditation on what remains true when our perspective changes.

### Phase 1: The Equation of the Rod

Let us first capture the essence of this rod in our original coordinate system. We know that a line with intercepts $a$ and $b$ is described by the elegant intercept form:

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

This equation is the 'DNA' of our line. It tells us everything we need to know about its position.

To make this useful for our distance calculations, we must transform it into the general form, $Ax + By + C = 0$. By rearranging, we get:

$$ \frac{1}{a}x + \frac{1}{b}y - 1 = 0 $$

Here, $A = \frac{1}{a}$, $B = \frac{1}{b}$, and $C = -1$. This is the algebraic soul of our line.

### Phase 2: The Invariant Truth

Now, we introduce the hero of our story: the perpendicular distance $d$ from the origin to the line. Why is this distance so important? Because it is an **invariant**.

Whether you look at the rod from the original axes or the rotated axes, the shortest distance from the origin to the rod is a physical property of the rod itself. It does not change.

The formula for the perpendicular distance from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by:

$$ d = \frac{|C|}{\sqrt{A^2 + B^2}} $$

Substituting our values, we find:

$$ d = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} $$

Simplifying this, we get:

$$ d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} $$

This is our first anchor point. Keep this equation safe; it represents the 'true' distance of the rod from the origin.

### Phase 3: The Illusion of Rotation

Now, we rotate the axes by an angle $	heta$. The diagram might look different, but remember: the rod has not moved. The origin has not moved.

Therefore, the distance $d$ must be identical. In the new coordinate system, the intercepts are $p$ and $q$.

Following the same logic, the equation of the line in this new frame is $\frac{x'}{p} + \frac{y'}{q} = 1$. The perpendicular distance $d$ in this new system is:

$$ d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}} $$

### Phase 4: The Grand Synthesis

We have two expressions for the same physical distance $d$. Since $d = d$, we can equate them:

$$ \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}} $$

To reveal the final relationship, we square both sides to eliminate the square roots:

$$ \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{1}{p^2} + \frac{1}{q^2}} $$

Taking the reciprocal of both sides, we arrive at the beautiful, symmetric result:

$$ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2} $$

This result is not just an answer to a multiple-choice question; it is a testament to the power of invariance. In physics and mathematics, finding quantities that do not change under transformation is the key to unlocking the deepest secrets of the universe. You have just performed a transformation, observed the invariance, and derived a fundamental truth. Well done!

Animated Solution for Mathematics - Straight Lines: Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ , then

Select Answer:

Visualized Solution

The Original Setup

Equation of Line $L$

Distance from the Origin

Distance Formula

Distance in Terms of $a$ and $b$

Simplifying the Distance

Rotating the Axes

New Intercepts $p$ and $q$

New Equation of Line $L$

Distance in Terms of $p$ and $q$

The Principle of Invariance

Equating the Distances

Squaring Both Sides

Final Relation

The Sigma Insight: Distance of a Point from a Line

The Geometry of Invariance

Phase 1

Phase 2

Phase 3

Phase 4

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