If the lines kx-2y-1=0 and 6x-4y-m=0 are identical (coindent) lines, then the values of k and m

Options

A. k=3, m=2
B. k=-3, m=2
C. k=-3, m=-2
D. k=3, m=-2

Explanation

To find the values of \( k \) and \( m \) such that the lines \( kx - 2y - 1 = 0 \) and \( 6x - 4y - m = 0 \) are identical (coincident), we can use the condition for two lines to be coincident.

### Step-by-Step Solution:

1. **Identify the coefficients**:
For the first line \( kx - 2y - 1 = 0 \), the coefficients are:
- \( A_1 = k \)
- \( B_1 = -2 \)
- \( C_1 = -1 \)

For the second line \( 6x - 4y - m = 0 \), the coefficients are:
- \( A_2 = 6 \)
- \( B_2 = -4 \)
- \( C_2 = -m \)

2. **Set up the condition for coincident lines**:
The lines are coincident if the ratios of their coefficients are equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]
This gives us three equations to work with:
\[ \frac{k}{6} = \frac{-2}{-4} = \frac{-1}{-m} \]

3. **Simplify the ratios**:
- From \( \frac{-2}{-4} \), we simplify to \( \frac{1}{2} \).
- Thus, we have:
\[ \frac{k}{6} = \frac{1}{2} \]
and
\[ \frac{-1}{-m} = \frac{1}{2} \]

4. **Solve for \( k \)**:
From \( \frac{k}{6} = \frac{1}{2} \):
\[ k = 6 \cdot \frac{1}{2} = 3 \]

5. **Solve for \( m \)**:
From \( \frac{-1}{-m} = \frac{1}{2} \):
\[ \frac{1}{m} = \frac{1}{2} \implies m = 2 \]

6. **Final values**:
Therefore, the values of \( k \) and \( m \) are:
\[ k = 3, \quad m = 2 \]

### Summary:
The values of \( k \) and \( m \) such that the lines are coincident are:
- \( k = 3 \)
- \( m = 2 \)