If the slope of one of the lines represented by `ax^(2)+(3a+1)xy+3y^(2)=0` be reciprocal of the slope of the other, then the slope of the lines are

Options

A. `3/2,2/3`
B. `1/2,2/1`
C. `1/3,3`
D. `(-1)/3,-3`

Explanation

To solve the problem, we need to find the slopes of the lines represented by the equation \( ax^2 + (3a + 1)xy + 3y^2 = 0 \) under the condition that the slopes are reciprocals of each other.

### Step-by-Step Solution

1. **Identify the coefficients**:
The given equation is \( ax^2 + (3a + 1)xy + 3y^2 = 0 \). Here, we can identify:
- \( a' = a \)
- \( 2h' = 3a + 1 \) (coefficient of \( xy \))
- \( b' = 3 \)

2. **Use the formulas for slopes**:
For a pair of straight lines represented by the general equation \( a'x^2 + 2h'xy + b'y^2 = 0 \):
- The sum of the slopes \( m_1 + m_2 = -\frac{2h'}{b'} \)
- The product of the slopes \( m_1 m_2 = \frac{a'}{b'} \)

3. **Calculate \( m_1 + m_2 \)**:
\[ m_1 + m_2 = -\frac{2(3a + 1)}{3} = -\frac{6a + 2}{3} \]

4. **Calculate \( m_1 m_2 \)**:
\[ m_1 m_2 = \frac{a}{3} \]

5. **Set up the relationship between slopes**:
Given that the slopes are reciprocals, we have:
\[ m_2 = \frac{1}{m_1} \implies m_1 m_2 = 1 \]
Therefore, we can set:
\[ m_1 m_2 = 1 = \frac{a}{3} \]
This gives us:
\[ a = 3 \]

6. **Substitute \( a \) back into the sum of slopes**:
Now substituting \( a = 3 \) into the equation for \( m_1 + m_2 \):
\[ m_1 + m_2 = -\frac{6(3) + 2}{3} = -\frac{18 + 2}{3} = -\frac{20}{3} \]

7. **Set up the quadratic equation**:
We now have:
\[ m_1 + m_2 = -\frac{20}{3} \quad \text{and} \quad m_1 m_2 = 1 \]
This leads to the quadratic equation:
\[ x^2 + \frac{20}{3}x + 1 = 0 \]

8. **Multiply through by 3 to eliminate the fraction**:
\[ 3x^2 + 20x + 3 = 0 \]

9. **Use the quadratic formula to find the slopes**:
The roots of the equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ = \frac{-20 \pm \sqrt{400 - 36}}{6} = \frac{-20 \pm \sqrt{364}}{6} = \frac{-20 \pm 2\sqrt{91}}{6} = \frac{-10 \pm \sqrt{91}}{3} \]

10. **Final slopes**:
Thus, the slopes of the lines are:
\[ m_1 = \frac{-10 + \sqrt{91}}{3}, \quad m_2 = \frac{-10 - \sqrt{91}}{3} \]