If the equation `x^(2)+3xy+2y^(2)+x-y+k=0` represents a pair of line, then k=

Options

A. `-12`
B. `-6`
C. `6`
D. `12`

Answer

To determine the value of \( k \) for which the equation \( x^2 + 3xy + 2y^2 + x - y + k = 0 \) represents a pair of straight lines, we will follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared with the general form of a conic section:
\[ ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \]
From the equation \( x^2 + 3xy + 2y^2 + x - y + k = 0 \), we identify:
- \( a = 1 \)
- \( b = 2 \)
- \( h = \frac{3}{2} \)
- \( g = \frac{1}{2} \)
- \( f = -\frac{1}{2} \)
- \( c = k \)

### Step 2: Set up the determinant condition
For the equation to represent a pair of straight lines, the determinant \( \Delta \) must be zero. The determinant is given by:
\[ \Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \]
Substituting the values we identified:
\[ \Delta = \begin{vmatrix} 1 & \frac{3}{2} & \frac{1}{2} \\ \frac{3}{2} & 2 & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & k \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ \Delta = 1 \left( 2k + \frac{3}{4} \right) - \frac{3}{2} \left( \frac{1}{2} k + \frac{1}{4} \right) + \frac{1}{2} \left( -\frac{3}{4} - 2 \right) \]
This simplifies to:
\[ \Delta = 2k + \frac{3}{4} - \frac{3}{4}k - \frac{3}{8} - \frac{3}{8} \]
Combining terms gives:
\[ \Delta = (2 - \frac{3}{4})k + \left( \frac{3}{4} - \frac{3}{8} - \frac{3}{8} \right) \]
\[ = \frac{5}{4}k + 0 \]

### Step 4: Set the determinant to zero
For the equation to represent a pair of lines, we set \( \Delta = 0 \):
\[ \frac{5}{4}k = 0 \]
Thus, solving for \( k \):
\[ k = 0 \]

### Step 5: Final answer
The value of \( k \) for which the equation represents a pair of straight lines is:
\[ \boxed{0} \]