1. Key Concepts and Formulas

• A homogeneous second-degree equation of the form $Ax^2 + Bxy + Cy^2 = 0$ represents a pair of straight lines passing through the origin.
• If the line $l_1x + m_1y = 0$ is one of the lines represented by the equation $Ax^2 + Bxy + Cy^2 = 0$, then substituting $y = -\frac{l_1}{m_1}x$ (or $x = -\frac{m_1}{l_1}y$) into the equation of the pair of lines will result in an equation that holds true for all x (or y), implying the coefficient of $x^2$ (or $y^2$) must be zero.
• Alternatively, if $y=mx$ is one of the lines represented by $Ax^2 + Bxy + Cy^2 = 0$, then $Cm^2 + Bm + A = 0$.

2. Step-by-Step Solution

We are given the equation of a pair of straight lines and the equation of one of the lines. Our goal is to find the value of the unknown constant $c$.

Step 1: Identify Given Information

• Equation of the pair of straight lines: $6x^2 - xy + 4cy^2 = 0$
• Equation of one of the lines: $3x + 4y = 0$

Step 2: Express one variable in terms of the other from the given line Since the line $3x + 4y = 0$ is one of the lines represented by the given pair, we can express one variable in terms of the other from this line's equation. This relationship must also hold true for the equation of the pair of lines. From $3x + 4y = 0$, we can write: $4y = -3x$ $y = -\frac{3}{4}x$ Explanation: We isolate $y$ to easily substitute it into the more complex equation of the pair of lines. This substitution simplifies the problem from two variables to a single variable.

Step 3: Substitute into the Pair of Lines Equation Now, substitute the expression for $y$ (from Step 2) into the equation of the pair of lines: $6x^2 - x\left(-\frac{3}{4}x\right) + 4c\left(-\frac{3}{4}x\right)^2 = 0$ Explanation: If $3x+4y=0$ is indeed one of the lines, then any point $(x,y)$ that satisfies $3x+4y=0$ must also satisfy $6x^2 - xy + 4cy^2 = 0$. By substituting $y = -\frac{3}{4}x$, we are essentially checking this condition.

Step 4: Simplify and Solve for the Unknown Coefficient Let's simplify the equation obtained in Step 3: $6x^2 + \frac{3}{4}x^2 + 4c\left(\frac{9}{16}x^2\right) = 0$ $6x^2 + \frac{3}{4}x^2 + \frac{9c}{4}x^2 = 0$ Now, we can factor out $x^2$ from all terms: $x^2 \left(6 + \frac{3}{4} + \frac{9c}{4}\right) = 0$ For this equation to hold true for all points $(x,y)$ on the line $3x+4y=0$ (other than the origin $(0,0)$ where $x=0$), the expression inside the parenthesis must be equal to zero. If $x=0$, then $y=0$, which corresponds to the origin (a point common to all lines represented by homogeneous equations). For any other point on the line, $x \neq 0$, so we can divide by $x^2$. Therefore, we must have: $6 + \frac{3}{4} + \frac{9c}{4} = 0$ To eliminate the fractions, multiply the entire equation by 4: $4(6) + 4\left(\frac{3}{4}\right) + 4\left(\frac{9c}{4}\right) = 0 \times 4$ $24 + 3 + 9c = 0$ $27 + 9c = 0$ $9c = -27$ $c = -\frac{27}{9}$ $c = -3$

Thus, the value of $c$ is $-3$.

3. Common Mistakes & Tips

• Understanding the "Why": The core idea is that if a line is a factor of a polynomial, then substituting the line's equation into the polynomial must make the polynomial identically zero. For a homogeneous equation like $x^2(\text{constant}) = 0$, this implies the constant must be zero (unless $x=0$, which is the trivial case of the origin).
• Alternative Method (Using Slopes): For a pair of lines $Ax^2 + Bxy + Cy^2 = 0$, if $y=mx$ is one of the lines, then $m$ must satisfy the quadratic equation $Cm^2 + Bm + A = 0$.
  • From $3x + 4y = 0$, the slope is $m = -\frac{3}{4}$.
  • Comparing $6x^2 - xy + 4cy^2 = 0$ with $Ax^2 + Bxy + Cy^2 = 0$, we have $A=6$, $B=-1$, $C=4c$.
  • Substitute these into $Cm^2 + Bm + A = 0$: $(4c)\left(-\frac{3}{4}\right)^2 + (-1)\left(-\frac{3}{4}\right) + 6 = 0$ $4c\left(\frac{9}{16}\right) + \frac{3}{4} + 6 = 0$ $\frac{9c}{4} + \frac{3}{4} + \frac{24}{4} = 0$ Multiply by 4: $9c + 3 + 24 = 0 \Rightarrow 9c + 27 = 0 \Rightarrow 9c = -27 \Rightarrow c = -3$. This method is often quicker and less prone to algebraic errors if you are comfortable with the slope transformation.
• Common Algebraic Mistakes: Be careful with signs, squaring fractions, and combining terms. Forgetting to multiply all terms by the common denominator when clearing fractions is a frequent error.

4. Summary

When one of the lines forming a pair of straight lines (represented by a homogeneous quadratic equation) is given, its equation can be used to find unknown coefficients in the pair's equation. This is done by substituting the relationship between $x$ and $y$ (derived from the given line) into the equation of the pair of lines. The resulting expression, when simplified, must be identically zero, allowing us to solve for the unknown. Using this method, we found that $c = -3$.

The final answer is $\boxed{-3}$. This corresponds to option (A).

5. Final Answer

The final answer is $\boxed{-3}$, which corresponds to option (A).