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

Options

A. `x-sqrt(3)y=0 and sqrt(3)x+y=0`
B. `x+sqrt(3)y=0 and sqrt(3)x-y=0`
C. `x-3y=0 and 3x+y=0`
D. `x+3y=0 and 3x+y=0`

Answer

To find the equations of the lines represented by the equation \(3x^2 - 2\sqrt{3}xy - 3y^2 = 0\), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 3x^2 - 2\sqrt{3}xy - 3y^2 = 0 \]

2. **Divide the entire equation by \(x^2\) (assuming \(x \neq 0\)):**
\[ 3 - \frac{2\sqrt{3}y}{x} - \frac{3y^2}{x^2} = 0 \]
This can be rewritten as:
\[ \frac{3y^2}{x^2} + \frac{2\sqrt{3}y}{x} - 3 = 0 \]

3. **Let \(y = mx\) (where \(m\) is the slope of the line):**
Substitute \(y = mx\) into the equation:
\[ 3m^2 + 2\sqrt{3}m - 3 = 0 \]

4. **This is a quadratic equation in \(m\):**
The standard form is \(am^2 + bm + c = 0\), where:
- \(a = 3\)
- \(b = 2\sqrt{3}\)
- \(c = -3\)

5. **Use the quadratic formula to find the roots:**
The quadratic formula is given by:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ m = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
\[ = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{6} \]
\[ = \frac{-2\sqrt{3} \pm \sqrt{48}}{6} \]
\[ = \frac{-2\sqrt{3} \pm 4\sqrt{3}}{6} \]

6. **Simplify the expression for \(m\):**
\[ m = \frac{2\sqrt{3}}{6} \quad \text{or} \quad m = \frac{-6\sqrt{3}}{6} \]
\[ = \frac{1}{\sqrt{3}} \quad \text{or} \quad -\sqrt{3} \]

7. **Write the equations of the lines:**
Using \(y = mx\):
- For \(m = \frac{1}{\sqrt{3}}\):
\[ y = \frac{1}{\sqrt{3}}x \quad \Rightarrow \quad \sqrt{3}y - x = 0 \]
- For \(m = -\sqrt{3}\):
\[ y = -\sqrt{3}x \quad \Rightarrow \quad \sqrt{3}x + y = 0 \]

### Final Equations:
The equations of the lines represented by the given quadratic equation are:
1. \(\sqrt{3}y - x = 0\)
2. \(\sqrt{3}x + y = 0\)