The combined equation of the two lines ax + by + c = 0 and a'x + b'y + c' = 0 can be written as (ax

Q: Wait, why is the equation $2x^2 + xy - 3y^2 = 0$ homogeneous? Does that matter?

That is a fantastic observation! A homogeneous equation of the second degree, where every term has a total degree of two (like $x^2$, $xy$, and $y^2$), always represents a pair of lines passing through the origin. This is a special property that allows us to use the elegant angle bisector formula. If there were constant terms or linear terms, the lines wouldn't pass through the origin, and we would need a much more complex approach.

Q: Can I just factorize the equation to find the two lines and then find their bisectors individually?

You absolutely can! Factorizing $2x^2 + xy - 3y^2 = 0$ gives $(2x + 3y)(x - y) = 0$, so the lines are $2x + 3y = 0$ and $x - y = 0$. You could then use the angle bisector formula for individual lines. However, in a JEE exam, time is your most precious resource. The combined formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ is a shortcut designed to save you those critical minutes. Mastering it is part of the 'JEE mindset'!

Q: Why is the formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$? Where does it come from?

The formula is derived from the geometric definition of an angle bisector: the locus of points equidistant from the two lines. If you take the two lines $L_1: m_1x + n_1y = 0$ and $L_2: m_2x + n_2y = 0$, the bisectors satisfy $\frac{L_1}{\sqrt{m_1^2 + n_1^2}} = \pm \frac{L_2}{\sqrt{m_2^2 + n_2^2}}$. When you multiply these together to get the combined equation, the algebra simplifies beautifully into the form $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. It's a classic result that turns a long geometric problem into a simple algebraic substitution.

Question

The combined equation of the two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ can be written as $(ax + by + c)(a'x + b'y + c') = 0$. The equation of the angle bisectors of the lines represented by the equation $2x^2 + xy - 3y^2 = 0$ is

Options:

(A) $3x^2 + 5xy + 2y^2 = 0$

(B) $x^2 - y^2 + 10xy = 0$

(C) $3x^2 + xy - 2y^2 = 0$

(D) $x^2 - y^2 - 10xy = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Mastering Pair of Lines

Welcome, future engineer! Today, we are diving into one of the most elegant corners of coordinate geometry: the pair of straight lines.

When you see an equation like $2x^2 + xy - 3y^2 = 0$, I want you to stop seeing it as just a collection of numbers and variables. Instead, visualize two distinct lines, slicing through the origin $(0,0)$, creating a beautiful, symmetric intersection.

Our mission is to find the equation of the angle bisectors of these lines. This is not just a calculation; it is a journey into the symmetry of the plane.

### Phase 1: The Anatomy of the Equation

Every homogeneous equation of the second degree, $ax^2 + 2hxy + by^2 = 0$, is a template for a pair of lines passing through the origin. When we look at our given equation, $2x^2 + xy - 3y^2 = 0$, we are essentially looking at a specific instance of this general form.

To unlock the secrets of this equation, we must first extract the coefficients. By comparing $2x^2 + xy - 3y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$, we identify our parameters:

- $a = 2$
- $2h = 1 \implies h = \frac{1}{2}$
- $b = -3$

These three values, $a$, $h$, and $b$, are the DNA of our pair of lines. They contain all the information about the slopes and the angles of the lines.

### Phase 2: The Magic of the Formula

Now, here is where the JEE magic happens. You could spend time finding the slopes of the individual lines, but we want efficiency. We use the standard formula for the combined equation of the angle bisectors:

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

This formula is a gift. It encapsulates the geometric condition that any point on an angle bisector is equidistant from the two lines.

It transforms a potentially tedious geometric derivation into a clean, algebraic substitution. I want you to memorize this formula, not just as a string of symbols, but as a shortcut to precision.

### Phase 3: The Execution

Let's perform the substitution with care. We have $a = 2$, $b = -3$, and $h = \frac{1}{2}$. Plugging these into our formula, we get:

$$\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$$

Focus on the denominator on the left-hand side. It is $2 - (-3)$, which simplifies to $2 + 3 = 5$. On the right-hand side, we have $xy$ divided by $\frac{1}{2}$, which is equivalent to multiplying by $2$.

So, our equation becomes:

$$\frac{x^2 - y^2}{5} = 2xy$$

Now, we cross-multiply by $5$ to clear the fraction:

$$x^2 - y^2 = 10xy$$

Finally, we rearrange the terms to match the standard form of a pair of lines:

$$x^2 - 10xy - y^2 = 0$$

### Conclusion: The Elegance of the Result

Look at that! We have arrived at the final equation, **$x^2 - 10xy - y^2 = 0$**. This result is not just an answer; it is a testament to the power of systematic thinking.

By identifying the structure of the equation and applying the right tool, we bypassed the complexity and found the solution with confidence. Keep this clarity of thought as you tackle more problems.

You are not just solving equations; you are mastering the language of the universe. Keep pushing, keep visualizing, and most importantly, keep falling in love with the process!

Select Answer:

Visualized Solution

The Given Pair of Lines

Comparing with General Form

Extracting Coefficients

Finding the value of $h$

The Angle Bisector Formula

Substituting Values

Simplifying the Denominator

Simplifying the Right Hand Side

Equating Both Sides

Final Rearrangement

The Sigma Insight: Angle Between Two Lines

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Conclusion

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