If the pair of straight lines x² - 2pxy - y² = 0 and x² - 2qxy - y² = 0 be such that each pair bisec

Q: Why can't we just find the slopes of the lines and bisect them manually?

You absolutely could! But finding the slopes involves solving a quadratic equation, which leads to messy radicals. The formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ is a powerful, elegant shortcut derived from that exact process. In JEE, efficiency is key!

Q: Wait, why do we compare coefficients? Can't we just solve for the lines?

Solving for the lines individually is possible but often leads to algebraic traps. Comparing coefficients is a standard, robust technique in coordinate geometry to show that two equations represent the same locus. It's cleaner and less prone to calculation errors.

Q: Does the order of $p$ and $q$ matter?

Not at all! The condition is symmetric. If you swap $p$ and $q$, the condition $pq = -1$ remains unchanged. It reflects the beautiful, mutual nature of the relationship between these two pairs of lines.

Question

If the pair of straight lines $x^2 - 2pxy - y^2 = 0$ and $x^2 - 2qxy - y^2 = 0$ be such that each pair bisects the angle between the other pair, then

Options:

(A) $pq = -1$

(B) $p = q$

(C) $p = -q$

(D) $pq = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two pairs of lines passing through the origin. These aren't just random lines; they are locked in a sophisticated geometric dance.

We are given two pairs:
$L_1: x^2 - 2pxy - y^2 = 0$
$L_2: x^2 - 2qxy - y^2 = 0$

The problem states that each pair bisects the angle between the other. This is a profound condition: if you were to draw the angle bisectors of the lines in $L_2$, you would find yourself tracing the lines in $L_1$. They are, in essence, reflections of each other's symmetry.

### The Toolkit: The Angle Bisector Formula

To solve this, we need a powerful tool from our geometry arsenal. For any pair of lines represented by the general equation $ax^2 + 2hxy + by^2 = 0$, the joint equation of their angle bisectors is given by the elegant formula:

$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

This formula is the key that unlocks the entire problem. It allows us to jump from the lines themselves to their bisectors without ever needing to calculate the individual slopes.

### The Calculation: Applying the Tool to $L_2$

Let's focus our attention on the second pair, $L_2: x^2 - 2qxy - y^2 = 0$. By comparing this to our standard form $ax^2 + 2hxy + by^2 = 0$, we identify the coefficients: $a = 1$, $b = -1$, and $2h = -2q$, which simplifies to $h = -q$.

Now, we substitute these values into our bisector formula:

$$\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-q}$$

Simplifying the denominator on the left, $1 - (-1)$ becomes $2$. Thus, we have:

$$\frac{x^2 - y^2}{2} = \frac{xy}{-q}$$

By cross-multiplying, we get $-q(x^2 - y^2) = 2xy$, which rearranges beautifully into:

$$qx^2 + 2xy - qy^2 = 0$$

This is the equation of the bisectors of $L_2$.

### The Comparison: The Final Symmetry

Here is the 'Aha!' moment. We know that the bisectors of $L_2$ are exactly the lines of $L_1$. Therefore, the equation $qx^2 + 2xy - qy^2 = 0$ must represent the same pair of lines as $x^2 - 2pxy - y^2 = 0$.

When two equations represent the same pair of lines, their coefficients must be proportional. We set up the ratios:

$$\frac{q}{1} = \frac{2}{-2p} = \frac{-q}{-1}$$

Looking at these ratios, we see that $\frac{q}{1}$ and $\frac{-q}{-1}$ are identical, both equal to $q$. We then equate the first two ratios:

$$q = \frac{2}{-2p}$$

The twos cancel out, leaving us with $q = -\frac{1}{p}$. Multiplying both sides by $p$, we arrive at the final, elegant condition:

**$pq = -1$**

This result is not just a number; it is the mathematical signature of the perfect, mutual symmetry between these two pairs of lines. You have successfully navigated the geometry, applied the formula, and uncovered the hidden relationship.

Animated Solution for Mathematics - Straight Lines: If the pair of straight lines $x^{2} - 2 p x y - y^{2} = 0$ and $x^{2} - 2 q x y - y^{2} = 0$ be such that each pair bisects the angle between the other pair, then

Select Answer:

Visualized Solution

The Two Pairs of Lines

The Mutual Bisection Condition

Formula for Angle Bisectors

Identifying Coefficients for $L_{2}$

Substituting into the Bisector Formula

Simplifying the Denominator

Rearranging to Standard Form

Equating Bisectors to $L_{1}$

Comparing Coefficients

Extracting the Relation

Final Result

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Toolkit

The Calculation

The Comparison

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