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

Options

A. -3
B. -1
C. 3
D. 1

Answer

To solve the problem, we need to determine the value of \( c \) such that one of the lines represented by the equation \( 6x^2 - xy + 4cy^2 = 0 \) is the same as the line given by \( 3x + 4y = 0 \).

### Step-by-Step Solution:

1. **Identify the given equations**:
- The first equation is \( 6x^2 - xy + 4cy^2 = 0 \).
- The second equation is \( 3x + 4y = 0 \).

2. **Find the slope of the line \( 3x + 4y = 0 \)**:
- Rearranging the equation gives us \( y = -\frac{3}{4}x \).
- Therefore, the slope \( m \) of this line is \( -\frac{3}{4} \).

3. **Use the general form of the equation of pair of straight lines**:
- The general form of the equation of a pair of straight lines is given by \( ax^2 + 2hxy + by^2 = 0 \).
- Here, \( a = 6 \), \( h = -\frac{1}{2} \), and \( b = 4c \).

4. **Calculate the sum of the slopes**:
- The sum of the slopes \( m_1 + m_2 \) can be expressed as:
\[ m_1 + m_2 = -\frac{2h}{b} = -\frac{-1}{4c} = \frac{1}{4c} \]
- Since one slope \( m_1 = -\frac{3}{4} \), we can write:
\[ -\frac{3}{4} + m_2 = \frac{1}{4c} \quad \text{(Equation 1)} \]

5. **Calculate the product of the slopes**:
- The product of the slopes \( m_1 \cdot m_2 \) can be expressed as:
\[ m_1 \cdot m_2 = \frac{a}{b} = \frac{6}{4c} \]
- Substituting \( m_1 = -\frac{3}{4} \):
\[ -\frac{3}{4} m_2 = \frac{6}{4c} \quad \text{(Equation 2)} \]

6. **Solve Equation 2 for \( m_2 \)**:
- Rearranging gives:
\[ m_2 = -\frac{6}{3c} = -\frac{2}{c} \]

7. **Substitute \( m_2 \) back into Equation 1**:
- Substitute \( m_2 = -\frac{2}{c} \) into Equation 1:
\[ -\frac{3}{4} - \frac{2}{c} = \frac{1}{4c} \]

8. **Clear the fractions by multiplying through by \( 4c \)**:
- This gives:
\[ -3c - 8 = 1 \]
- Rearranging:
\[ -3c = 9 \quad \Rightarrow \quad c = -3 \]

### Final Answer:
Thus, the value of \( c \) is \( -3 \).