If the lines represented by `kx^(2) - 3xy + 6y^(2) = 0` are perpendicular to each other then…..

Options

A. k = 6
B. k = -6
C. k = 3
D. k = -4

Answer

To determine the value of \( k \) such that the lines represented by the equation \( kx^2 - 3xy + 6y^2 = 0 \) are perpendicular to each other, we can follow these steps:

### Step 1: Identify the coefficients
The general form of a pair of straight lines is given by:
\[ ax^2 + 2hxy + by^2 = 0 \]
From the given equation \( kx^2 - 3xy + 6y^2 = 0 \), we can identify:
- \( a = k \)
- \( 2h = -3 \) (thus \( h = -\frac{3}{2} \))
- \( b = 6 \)

### Step 2: Use the condition for perpendicular lines
For two lines represented by the equation to be perpendicular, the condition is:
\[ a + b = 0 \]
Substituting the values we identified:
\[ k + 6 = 0 \]

### Step 3: Solve for \( k \)
Now, we solve the equation:
\[ k + 6 = 0 \implies k = -6 \]

### Conclusion
Thus, the value of \( k \) for which the lines are perpendicular is:
\[ \boxed{-6} \]