The lines represented by the equation
`x^2 + 2sqrt(3)xy + 3y^(2) -3x -3sqrt(3)y -4=0`, are

Options

A. perpendicular to each other
B. parallel
C. inclined at `45^(@)` to each other
D. None of these

Answer

To solve the problem of determining the nature of the lines represented by the equation

\[ x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0, \]

we will follow these steps:

### Step 1: Identify coefficients
The general form of a second-degree equation representing two straight lines is given by:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0. \]

From the given equation, we can identify the coefficients as follows:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( h = \sqrt{3} \) (since \( 2h = 2\sqrt{3} \) implies \( h = \sqrt{3} \))
- \( b = 3 \) (coefficient of \( y^2 \))
- \( g = -\frac{3}{2} \) (since \( 2g = -3 \) implies \( g = -\frac{3}{2} \))
- \( f = -\frac{3\sqrt{3}}{2} \) (since \( 2f = -3\sqrt{3} \) implies \( f = -\frac{3\sqrt{3}}{2} \))
- \( c = -4 \) (constant term)

### Step 2: Calculate \( h^2 - ab \)
Next, we will calculate \( h^2 - ab \):

\[ h^2 = (\sqrt{3})^2 = 3, \]
\[ ab = 1 \cdot 3 = 3. \]

Thus,

\[ h^2 - ab = 3 - 3 = 0. \]

### Step 3: Determine the nature of the lines
The condition for the lines to be parallel is when \( h^2 - ab = 0 \). Since we found that \( h^2 - ab = 0 \), it indicates that the lines represented by the equation are indeed parallel.

### Step 4: Conclusion
Therefore, the lines represented by the given equation are parallel.

### Final Answer
The lines are parallel.

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