If graph of `x+2y+3=0` is perpendicular to the graph of `ax+3y+2=0`, then a equals

Options

A. `-6`
B. `-(3)/(2)`
C. `(2)/(3)`
D. `(3)/(2)`

Explanation

To solve the problem, we need to find the value of \( a \) such that the lines represented by the equations \( x + 2y + 3 = 0 \) and \( ax + 3y + 2 = 0 \) are perpendicular to each other.

### Step-by-step Solution:

1. **Identify the equations of the lines:**
- The first line is given by \( x + 2y + 3 = 0 \).
- The second line is given by \( ax + 3y + 2 = 0 \).

2. **Convert the first line to slope-intercept form (y = mx + c):**
- Start with the equation:
\[ x + 2y + 3 = 0 \]
- Rearranging gives:
\[ 2y = -x - 3 \]
- Dividing by 2:
\[ y = -\frac{1}{2}x - \frac{3}{2} \]
- The slope \( m_1 \) of the first line is \( -\frac{1}{2} \).

3. **Convert the second line to slope-intercept form:**
- Start with the equation:
\[ ax + 3y + 2 = 0 \]
- Rearranging gives:
\[ 3y = -ax - 2 \]
- Dividing by 3:
\[ y = -\frac{a}{3}x - \frac{2}{3} \]
- The slope \( m_2 \) of the second line is \( -\frac{a}{3} \).

4. **Use the property of perpendicular lines:**
- For two lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
- Substituting the slopes:
\[ \left(-\frac{1}{2}\right) \cdot \left(-\frac{a}{3}\right) = -1 \]
- This simplifies to:
\[ \frac{a}{6} = -1 \]

5. **Solve for \( a \):**
- Multiply both sides by 6:
\[ a = -6 \]

### Final Answer:
The value of \( a \) is \( -6 \).

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