`a(x^(2)-y^(2))+xy=0` represents a pair of straight lines for

Options

A. `a=1` only
B. `a=1` or `-1` only
C. `a=0` only
D. all real values of a.

Explanation

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

### Step 1: Rewrite the equation
The given equation is:
\[ a(x^2 - y^2) + xy = 0 \]
This can be rewritten as:
\[ ax^2 - ay^2 + xy = 0 \]

### Step 2: Identify coefficients
We can identify the coefficients from the standard form of a second-degree equation:
\[ Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0 \]
From our equation, we have:
- \( A = a \)
- \( B = -a \)
- \( 2H = 1 \) (thus \( H = \frac{1}{2} \))
- \( G = 0 \)
- \( F = 0 \)
- \( C = 0 \)

### Step 3: Set up the determinant condition
For the equation to represent a pair of straight lines, the determinant (denoted as \( \Delta \)) must be equal to zero. The determinant is given by:
\[ \Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 \]
Substituting the values we have:
\[ \Delta = a(-a)(0) + 2(0)(0)(\frac{1}{2}) - a(0)^2 - (-a)(0)^2 - (0)(\frac{1}{2})^2 \]
This simplifies to:
\[ \Delta = 0 + 0 - 0 - 0 - 0 = 0 \]

### Step 4: Solve the determinant condition
Since \( \Delta = 0 \), we need to check the condition for \( a \):
- The equation will represent a pair of straight lines if \( a \neq 0 \) (as \( a = 0 \) would make the equation linear).

### Conclusion
Thus, the equation \( a(x^2 - y^2) + xy = 0 \) represents a pair of straight lines for all values of \( a \) except \( a = 0 \).