If `x^(2)-kxy+y^(2)+2y+2=0` denotes a pair of straight lines then k =

Options

A. 2
B. `1//sqrt2`
C. `2sqrt2`
D. `sqrt2`

Answer

To solve the problem, we need to determine the value of \( k \) such that the equation \( x^2 - kxy + y^2 + 2y + 2 = 0 \) represents a pair of straight lines. We will use the condition that the determinant of the coefficients must equal zero.

### Step-by-Step Solution:

1. **Identify coefficients**:
The given equation is:
\[ x^2 - kxy + y^2 + 2y + 2 = 0 \]
We can rewrite this in the standard form:
\[ ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \]
From the equation, we identify:
- \( a = 1 \)
- \( b = 1 \)
- \( h = -\frac{k}{2} \)
- \( g = 0 \)
- \( f = 1 \)
- \( c = 2 \)

2. **Set up the determinant**:
The condition for the equation to represent a pair of straight lines is given by the determinant:
\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]
Substituting the identified values:
\[ \begin{vmatrix} 1 & -\frac{k}{2} & 0 \\ -\frac{k}{2} & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
We can calculate the determinant using the formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Substituting our values:
\[ D = 1 \left( 1 \cdot 2 - 1 \cdot 0 \right) - \left(-\frac{k}{2}\right) \left(0 - 1 \cdot 0\right) + 0 \]
Simplifying this gives:
\[ D = 2 - \left(-\frac{k}{2}\right)(1) = 2 + \frac{k}{2} \]

4. **Set the determinant to zero**:
For the equation to represent a pair of straight lines, we set the determinant to zero:
\[ 2 + \frac{k}{2} = 0 \]

5. **Solve for \( k \)**:
Rearranging gives:
\[ \frac{k}{2} = -2 \implies k = -4 \]

6. **Check for the condition**:
The value of \( k \) must satisfy the condition \( 2 - k^2 = 0 \):
\[ 2 - k^2 = 0 \implies k^2 = 2 \implies k = \pm \sqrt{2} \]

### Final Result:
Thus, the values of \( k \) that satisfy the condition are:
\[ k = \sqrt{2} \quad \text{or} \quad k = -\sqrt{2} \]