If the sum of the slopes of the lines given by `2x^(2)+kxy-3y^(2)=0` is equal to their product, then k=

Options

A. `-2`
B. `2`
C. `(-2)/(3)`
D. `(-2)/(9)`

Answer

To solve the problem, we need to find the value of \( k \) such that the sum of the slopes of the lines represented by the equation \( 2x^2 + kxy - 3y^2 = 0 \) is equal to their product.

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ 2x^2 + kxy - 3y^2 = 0 \]

2. **Divide the entire equation by \( x^2 \)**:
\[ 2 + \frac{ky}{x} - 3\left(\frac{y}{x}\right)^2 = 0 \]
Let \( m = \frac{y}{x} \) (where \( m \) is the slope of the line). The equation becomes:
\[ 2 + km - 3m^2 = 0 \]

3. **Rearrange the equation**:
\[ -3m^2 + km + 2 = 0 \]
This is a quadratic equation in terms of \( m \).

4. **Identify the coefficients**:
The quadratic equation can be expressed as:
\[ Am^2 + Bm + C = 0 \]
where \( A = -3 \), \( B = k \), and \( C = 2 \).

5. **Use the formulas for the sum and product of the roots**:
- The sum of the roots \( m_1 + m_2 \) is given by:
\[ m_1 + m_2 = -\frac{B}{A} = -\frac{k}{-3} = \frac{k}{3} \]
- The product of the roots \( m_1 m_2 \) is given by:
\[ m_1 m_2 = \frac{C}{A} = \frac{2}{-3} = -\frac{2}{3} \]

6. **Set the sum equal to the product**:
According to the problem, the sum of the slopes is equal to their product:
\[ \frac{k}{3} = -\frac{2}{3} \]

7. **Solve for \( k \)**:
Multiply both sides by 3:
\[ k = -2 \]

### Final Answer:
The value of \( k \) is \( -2 \).