Find the value of k for which the lines `5x+3y+2=0` and `3x-ky+6=0` are perpendicular.

Options

A. 5
B. 4
C. 3
D. 2

Answer

To find the value of \( k \) for which the lines \( 5x + 3y + 2 = 0 \) and \( 3x - ky + 6 = 0 \) are perpendicular, we can follow these steps:

### Step 1: Determine the slopes of the lines
The general form of a line is given by \( Ax + By + C = 0 \). The slope \( m \) of the line can be calculated using the formula:
\[ m = -\frac{A}{B} \]

For the first line \( 5x + 3y + 2 = 0 \):
- Here, \( A = 5 \) and \( B = 3 \).
- The slope \( m_1 \) is:
\[ m_1 = -\frac{5}{3} \]

For the second line \( 3x - ky + 6 = 0 \):
- Here, \( A = 3 \) and \( B = -k \).
- The slope \( m_2 \) is:
\[ m_2 = -\frac{3}{-k} = \frac{3}{k} \]

### Step 2: Use the condition for perpendicular lines
Two lines are perpendicular if the product of their slopes is equal to \(-1\):
\[ m_1 \cdot m_2 = -1 \]

Substituting the slopes we found:
\[ \left(-\frac{5}{3}\right) \cdot \left(\frac{3}{k}\right) = -1 \]

### Step 3: Simplify the equation
Simplifying the left-hand side:
\[ -\frac{5 \cdot 3}{3 \cdot k} = -1 \]
The \( 3 \) in the numerator and denominator cancels out:
\[ -\frac{5}{k} = -1 \]

### Step 4: Solve for \( k \)
Removing the negative signs from both sides gives:
\[ \frac{5}{k} = 1 \]
Cross-multiplying yields:
\[ 5 = k \]

### Conclusion
The value of \( k \) for which the lines are perpendicular is:
\[ \boxed{5} \]