If the equation `ax^(2)+2hxy+by^(2)=0` represented two lines `y=m_(1)x` and `y=m_(2)x` the

Options

A. `m_(1)+m_(2)=(-2h)/b` and `m_(1)m_(2)=a/b`
B. `m_(1)+m_(2)=(2h)/b` and `m_(1)m_(2)=(-a)/b`
C. `m_(1)+m_(2)=(2h)/b` and `m_(1)m_(2)=a/b`
D. `m_(1)+m_(2)=(2h)/b` and `m_(1)m_(2)=-ab`

Explanation

To solve the problem, we need to find the relationship between the slopes \( m_1 \) and \( m_2 \) of the two lines represented by the equation \( ax^2 + 2hxy + by^2 = 0 \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( ax^2 + 2hxy + by^2 = 0 \) represents a pair of straight lines. We can express these lines in the form \( y = m_1 x \) and \( y = m_2 x \).

2. **Forming the Joint Equation**:
The joint equation of the two lines can be expressed as:
\[ (y - m_1 x)(y - m_2 x) = 0 \]
Expanding this, we get:
\[ y^2 - (m_1 + m_2)xy + m_1 m_2 x^2 = 0 \]

3. **Comparing Coefficients**:
Now, we can compare the coefficients of the expanded equation with the original equation \( ax^2 + 2hxy + by^2 = 0 \):
- Coefficient of \( x^2 \): \( m_1 m_2 = \frac{a}{b} \)
- Coefficient of \( xy \): \( -(m_1 + m_2) = \frac{2h}{b} \)
- Coefficient of \( y^2 \): \( 1 = \frac{b}{b} \)

4. **Finding Relations**:
From the coefficient of \( x^2 \):
\[ m_1 m_2 = \frac{a}{b} \tag{1} \]
From the coefficient of \( xy \):
\[ m_1 + m_2 = -\frac{2h}{b} \tag{2} \]

5. **Final Relations**:
We have derived two important relationships:
- The product of the slopes:
\[ m_1 m_2 = \frac{a}{b} \]
- The sum of the slopes:
\[ m_1 + m_2 = -\frac{2h}{b} \]

### Summary of Results:
- The relationship between the slopes \( m_1 \) and \( m_2 \) is given by:
\[ m_1 m_2 = \frac{a}{b} \quad \text{and} \quad m_1 + m_2 = -\frac{2h}{b} \]