If the equation `2x^(2)-3xy+y^(2)-kx+5y+6=0` represents a pair of line, then k=

Options

A. `7, 8`
B. `-7, -8`
C. `-7, 8`
D. `7, -8`

Answer

To determine the value of \( k \) such that the equation

\[ 2x^2 - 3xy + y^2 - kx + 5y + 6 = 0 \]

represents a pair of straight lines, we will use the condition that the determinant \( \Delta \) must equal zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The general form of the equation of a pair of straight lines is given by:
\[ ax^2 + bxy + cy^2 + 2gx + 2fy + h = 0 \]
From the given equation, we can identify:
- \( a = 2 \)
- \( b = -3 \)
- \( c = 1 \)
- \( g = -\frac{k}{2} \)
- \( f = \frac{5}{2} \)
- \( h = 6 \)

2. **Set up the determinant**:
The determinant \( \Delta \) is given by:
\[ \Delta = \begin{vmatrix} a & \frac{b}{2} & g \\ \frac{b}{2} & c & f \\ g & f & h \end{vmatrix} \]
Plugging in the values we have:
\[ \Delta = \begin{vmatrix} 2 & -\frac{3}{2} & -\frac{k}{2} \\ -\frac{3}{2} & 1 & \frac{5}{2} \\ -\frac{k}{2} & \frac{5}{2} & 6 \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant, we have:
\[ \Delta = 2 \left( 1 \cdot 6 - \frac{5}{2} \cdot \frac{5}{2} \right) - \left( -\frac{3}{2} \right) \left( -\frac{3}{2} \cdot 6 - \frac{5}{2} \cdot -\frac{k}{2} \right) - \left( -\frac{k}{2} \right) \left( -\frac{3}{2} \cdot \frac{5}{2} - 1 \cdot -\frac{k}{2} \right) \]

Simplifying this will yield a quadratic equation in \( k \).

4. **Set the determinant to zero**:
For the equation to represent a pair of lines, set \( \Delta = 0 \). This will give us a quadratic equation in \( k \).

5. **Solve for \( k \)**:
After simplifying the determinant and setting it to zero, we will solve the resulting quadratic equation for \( k \).

6. **Final values of \( k \)**:
The solutions will yield the values of \( k \) that satisfy the condition for the equation to represent a pair of lines.

### Final Answer:
After performing the calculations, we find that \( k = 8 \) and \( k = 7 \).