If one of the lines given by 6x² - xy + 4cy² = 0 is 3x + 4y = 0, then c equals (a) -3 (b) -1 (c) 3 (

Q: Wait, why can't we just compare coefficients with $(ax + by)(lx + my) = 0$?

You absolutely can! That is a very elegant method. However, expanding $(ax + by)(lx + my)$ and equating coefficients involves solving a system of equations. Substitution is often faster and less prone to algebraic errors when you are given one of the lines explicitly. Both paths lead to the same truth, but substitution is the direct route here.

Q: Why does the $x^2$ term cancel out? Does that mean $x$ is zero?

Great observation! We aren't saying $x$ is zero. We are saying that for the equation to be true for *every* point on the line (where $x$ can be any value), the coefficient of $x^2$ must vanish. If the coefficient wasn't zero, the equation would only be true for specific points, not the entire line. By setting the coefficient to zero, we ensure the line satisfies the equation for all $x$.

Q: Is there a geometric way to see why $c = -3$?

Yes! The joint equation $6x^2 - xy + 4cy^2 = 0$ represents two lines passing through the origin. By finding $c = -3$, we essentially 'tune' the equation so that the two lines it describes are exactly the ones we need. It's like adjusting the focus on a telescope until the two lines snap into clear view.

Question

If one of the lines given by $6x^2 - xy + 4cy^2 = 0$ is $3x + 4y = 0$, then $c$ equals

Options:

(A) -3

(B) -1

(C) 3

(D) 1

Sigma Percentile · Accepted Answer

### The Geometry of Intersecting Lines: A Journey into Homogeneous Equations

My dear students, welcome to a beautiful exploration of coordinate geometry. Today, we are not just solving for a variable; we are uncovering the hidden structure of a pair of lines.

When you look at the equation $6x^2 - xy + 4cy^2 = 0$, I want you to see more than just symbols. I want you to see the 'DNA' of two straight lines that intersect at the origin. This is a homogeneous equation of the second degree, and it is a fundamental building block in our JEE journey.

### The Power of the Substitution Principle

We are given a specific line, $3x + 4y = 0$, and we are told that this line is a part of the pair represented by our quadratic equation. Think of the joint equation as a product of two linear factors.

If one factor is $3x + 4y = 0$, then every single point $(x, y)$ that lies on this line must also satisfy the joint equation. This is the Substitution Principle, and it is your most powerful tool in this scenario. We don't need to guess; we simply need to enforce the condition that the line belongs to the pair.

Let us isolate $y$ from our known line. By rearranging $3x + 4y = 0$, we get $4y = -3x$, which simplifies to $y = -\frac{3}{4}x$.

Now, we take this relationship and substitute it directly into our joint equation:

$$6x^2 - x\left(-\frac{3}{4}x\right) + 4c\left(-\frac{3}{4}x\right)^2 = 0$$

### The Algebraic Grind

I know that dealing with fractions like $-\frac{3}{4}$ can feel tedious, but stay with me. Precision here is the difference between a correct answer and a silly mistake.

Let's expand this carefully. The first term is $6x^2$. The middle term, $-x$ multiplied by $-\frac{3}{4}x$, becomes $+\frac{3}{4}x^2$.

Now, for the final term: $4c$ multiplied by the square of $-\frac{3}{4}x$. Squaring $-\frac{3}{4}$ gives us $\frac{9}{16}$, so we have:

$$4c \cdot \left(\frac{9}{16}\right)x^2 = \frac{9c}{4}x^2$$

Our equation now looks like this:

$$6x^2 + \frac{3}{4}x^2 + \frac{9c}{4}x^2 = 0$$

Notice something magical? Every term contains an $x^2$. We can factor it out:

$$x^2 \left(6 + \frac{3}{4} + \frac{9c}{4}\right) = 0$$

For this to hold true for all points on the line (where $x$ is not just zero), the expression inside the bracket must be zero. We have reduced a complex geometry problem into a simple linear equation:

$$6 + \frac{3}{4} + \frac{9c}{4} = 0$$

### The Final Victory

To clear the denominators, we multiply the entire equation by $4$. This gives us:

$$24 + 3 + 9c = 0$$

Combining our constants, we get $27 + 9c = 0$. A quick shift of the $27$ to the other side gives $9c = -27$, and dividing by $9$ yields our result:

**$c = -3$**

We have done it! By substituting the line into the joint equation, we have unlocked the value of $c$.

If you were to plug $c = -3$ back into the original equation, you would get $6x^2 - xy - 12y^2 = 0$. If you factorize this, you will find the two lines are indeed $3x + 4y = 0$ and $2x - 3y = 0$. This is the elegance of mathematics—everything fits together perfectly.

Animated Solution for Mathematics - Straight Lines: If one of the lines given by $6 x^{2} - x y + 4 c y^{2} = 0$ is $3 x + 4 y = 0$ , then $c$ equals

Select Answer:

Visualized Solution

Visualizing the Pair of Lines

Isolating $y$

The Substitution Principle

Raw Substitution

Expanding the Products

Simplifying the $c$ Term

Factoring out $x^{2}$

Clearing the Denominators

Solving for $c$

Conclusion and Takeaway

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Intersecting Lines

The Power of the Substitution Principle

The Algebraic Grind

The Final Victory

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the lines given by 6x2−xy+4cy2=0 is 3x+4y=0, then c equals