If `6x^2+11xy-10y^2+x+31y+k=0` represents a pair of straight lines , then k is equal to.

Options

A. `-15`
B. 6
C. `-10`
D. `-4`

Explanation

To determine the value of \( k \) such that the equation \( 6x^2 + 11xy - 10y^2 + x + 31y + k = 0 \) represents a pair of straight lines, we will use the condition that the determinant of the coefficients must be zero.

### Step-by-Step Solution:

1. **Identify Coefficients**:
The given equation can be expressed in the form \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \), where:
- \( A = 6 \)
- \( B = 11 \)
- \( C = -10 \)
- \( D = 1 \)
- \( E = 31 \)
- \( F = k \)

2. **Set Up the Determinant**:
For the equation to represent a pair of straight lines, the determinant must be zero:
\[ \begin{vmatrix} A & \frac{B}{2} & \frac{D}{2} \\ \frac{B}{2} & C & \frac{E}{2} \\ \frac{D}{2} & \frac{E}{2} & F \end{vmatrix} = 0 \]

Substituting the values of \( A, B, C, D, E, F \):
\[ \begin{vmatrix} 6 & \frac{11}{2} & \frac{1}{2} \\ \frac{11}{2} & -10 & \frac{31}{2} \\ \frac{1}{2} & \frac{31}{2} & k \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
The determinant can be calculated using the formula for a \( 3 \times 3 \) matrix:
\[ \text{Det} = A \begin{vmatrix} C & \frac{E}{2} \\ \frac{E}{2} & F \end{vmatrix} - \frac{B}{2} \begin{vmatrix} \frac{B}{2} & \frac{D}{2} \\ \frac{D}{2} & F \end{vmatrix} + \frac{D}{2} \begin{vmatrix} \frac{B}{2} & C \\ \frac{D}{2} & \frac{E}{2} \end{vmatrix} \]

Calculate each of the \( 2 \times 2 \) determinants:
- For \( A \):
\[ \begin{vmatrix} -10 & \frac{31}{2} \\ \frac{31}{2} & k \end{vmatrix} = (-10)k - \left(\frac{31}{2}\right)^2 = -10k - \frac{961}{4} \]

- For \( B \):
\[ \begin{vmatrix} \frac{11}{2} & \frac{1}{2} \\ \frac{1}{2} & k \end{vmatrix} = \left(\frac{11}{2}\right)k - \frac{1}{4} = \frac{11k}{2} - \frac{1}{4} \]

- For \( D \):
\[ \begin{vmatrix} \frac{11}{2} & -10 \\ \frac{1}{2} & \frac{31}{2} \end{vmatrix} = \left(\frac{11}{2}\right)\left(\frac{31}{2}\right) - (-10)\left(\frac{1}{2}\right) = \frac{341}{4} + 5 = \frac{341 + 20}{4} = \frac{361}{4} \]

4. **Combine Determinants**:
Substitute back into the determinant expression:
\[ 6\left(-10k - \frac{961}{4}\right) - \frac{11}{2}\left(\frac{11k}{2} - \frac{1}{4}\right) + \frac{1}{2}\left(\frac{361}{4}\right) = 0 \]

Simplifying gives:
\[ -60k - \frac{5766}{4} - \frac{121k}{4} + \frac{11}{8} + \frac{361}{8} = 0 \]

5. **Solve for \( k \)**:
Combine like terms and solve for \( k \):
\[ -\frac{240k + 121k}{4} = \frac{5766 - 361 - 11}{8} \]
\[ -\frac{361k}{4} = \frac{5766 - 372}{8} \]
\[ -\frac{361k}{4} = \frac{5394}{8} \]
\[ k = -\frac{5394 \times 4}{8 \times 361} \]
Simplifying gives \( k = -15 \).

### Final Answer:
Thus, the value of \( k \) is \( -15 \).