If the equation `3x^(2)+xy-y^(2)-3x+6y+k=0` represents a pair of straight lines, then the value of k, is

Options

A. 9
B. 1
C. `-9`
D. 0

Answer

To find the value of \( k \) such that the equation \( 3x^2 + xy - y^2 - 3x + 6y + k = 0 \) represents a pair of straight lines, we will follow these steps:

### Step 1: Identify coefficients
The general form of the equation representing a pair of straight lines is:
\[ ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \]
From the given equation \( 3x^2 + xy - y^2 - 3x + 6y + k = 0 \), we can identify the coefficients as follows:
- \( a = 3 \)
- \( b = -1 \)
- \( h = \frac{1}{2} \) (since the coefficient of \( xy \) is \( 1 \))
- \( g = -\frac{3}{2} \) (since the coefficient of \( x \) is \( -3 \))
- \( f = 3 \) (since the coefficient of \( y \) is \( 6 \), we take half for \( 2f \))
- \( c = k \)

### Step 2: Set up the determinant condition
For the equation to represent a pair of straight lines, the determinant of the matrix formed by these coefficients must equal zero:
\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]
Substituting the values we found:
\[ \begin{vmatrix} 3 & \frac{1}{2} & -\frac{3}{2} \\ \frac{1}{2} & -1 & 3 \\ -\frac{3}{2} & 3 & k \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand it:
\[ = 3 \begin{vmatrix} -1 & 3 \\ 3 & k \end{vmatrix} - \frac{1}{2} \begin{vmatrix} \frac{1}{2} & 3 \\ -\frac{3}{2} & k \end{vmatrix} - \left(-\frac{3}{2}\right) \begin{vmatrix} \frac{1}{2} & -1 \\ -\frac{3}{2} & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -1 & 3 \\ 3 & k \end{vmatrix} = -1 \cdot k - 3 \cdot 3 = -k - 9 \)

2. \( \begin{vmatrix} \frac{1}{2} & 3 \\ -\frac{3}{2} & k \end{vmatrix} = \frac{1}{2} \cdot k - 3 \cdot \left(-\frac{3}{2}\right) = \frac{k}{2} + \frac{9}{2} \)

3. \( \begin{vmatrix} \frac{1}{2} & -1 \\ -\frac{3}{2} & 3 \end{vmatrix} = \frac{1}{2} \cdot 3 - (-1) \cdot \left(-\frac{3}{2}\right) = \frac{3}{2} - \frac{3}{2} = 0 \)

Substituting these back into the determinant:
\[ 3(-k - 9) - \frac{1}{2}\left(\frac{k}{2} + \frac{9}{2}\right) + \frac{3}{2}(0) = 0 \]
This simplifies to:
\[ -3k - 27 - \frac{1}{4}k - \frac{9}{4} = 0 \]

### Step 4: Combine and solve for \( k \)
Combining the terms:
\[ -\left(3 + \frac{1}{4}\right)k - 27 - \frac{9}{4} = 0 \]
Finding a common denominator:
\[ -\frac{12k + k}{4} - \frac{108 + 9}{4} = 0 \]
This simplifies to:
\[ -\frac{13k + 117}{4} = 0 \]
Thus, we have:
\[ 13k + 117 = 0 \implies 13k = -117 \implies k = -9 \]

### Conclusion
The value of \( k \) is \( -9 \).