` 2x - ky +7=0 ` and ` 6x - 12 y +15 =0` has no solution for

Options

A. `k=-1`
B. `k=-4`
C. `k=4`
D. `k=1`

Answer

To find the value of \( k \) for which the lines represented by the equations \( 2x - ky + 7 = 0 \) and \( 6x - 12y + 15 = 0 \) have no solution, we will follow these steps:

### Step 1: Identify the coefficients
The given equations can be rewritten in the standard form \( Ax + By + C = 0 \).

For the first equation:
- \( A_1 = 2 \)
- \( B_1 = -k \)
- \( C_1 = 7 \)

For the second equation:
- \( A_2 = 6 \)
- \( B_2 = -12 \)
- \( C_2 = 15 \)

### Step 2: Apply the condition for no solution
Two lines have no solution if they are parallel, which occurs when:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]

### Step 3: Set up the ratio for the coefficients
Using the coefficients identified:
\[ \frac{2}{6} = \frac{-k}{-12} \]

### Step 4: Simplify the ratios
Simplifying \( \frac{2}{6} \) gives:
\[ \frac{1}{3} = \frac{k}{12} \]

### Step 5: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 1 \cdot 12 = 3k \]
\[ 12 = 3k \]

### Step 6: Solve for \( k \)
Dividing both sides by 3:
\[ k = \frac{12}{3} = 4 \]

### Conclusion
Thus, the value of \( k \) for which the two lines have no solution is:
\[ \boxed{4} \]