If the area of the parallelogram formed by the lines 2x - 3y + a = 0 , 3x - 2y - a = 0 , 2x - 3y + 3a = 0 and 3x - 2y - 2a = 0 is 10 square units , then a =

Options

A. `pm1`
B. `pm` 10
C. `pm` 5
D. none of these

Explanation

To solve the problem, we need to find the value of \( a \) such that the area of the parallelogram formed by the given lines is 10 square units.

### Step-by-Step Solution:

1. **Identify the Lines**:
The lines given are:
- Line 1: \( 2x - 3y + a = 0 \)
- Line 2: \( 3x - 2y - a = 0 \)
- Line 3: \( 2x - 3y + 3a = 0 \)
- Line 4: \( 3x - 2y - 2a = 0 \)

2. **Rewrite the Lines in Slope-Intercept Form**:
We need to express these lines in the form \( y = mx + c \) to identify their slopes and intercepts.

- For Line 1:
\[ 2x - 3y + a = 0 \implies 3y = 2x + a \implies y = \frac{2}{3}x + \frac{a}{3} \]
Here, the slope \( m_1 = \frac{2}{3} \) and intercept \( c_1 = \frac{a}{3} \).

- For Line 2:
\[ 3x - 2y - a = 0 \implies 2y = 3x - a \implies y = \frac{3}{2}x - \frac{a}{2} \]
Here, the slope \( m_2 = \frac{3}{2} \) and intercept \( d_1 = -\frac{a}{2} \).

- For Line 3:
\[ 2x - 3y + 3a = 0 \implies 3y = 2x + 3a \implies y = \frac{2}{3}x + a \]
Here, the slope \( m_3 = \frac{2}{3} \) and intercept \( c_2 = a \).

- For Line 4:
\[ 3x - 2y - 2a = 0 \implies 2y = 3x - 2a \implies y = \frac{3}{2}x - a \]
Here, the slope \( m_4 = \frac{3}{2} \) and intercept \( d_2 = -a \).

3. **Calculate the Area of the Parallelogram**:
The area \( A \) of the parallelogram formed by two pairs of lines can be calculated using the formula:
\[ A = \frac{|c_1 - c_2| \cdot |d_1 - d_2|}{|m_1 - m_2|} \]
Substituting the values:
\[ A = \frac{\left|\frac{a}{3} - a\right| \cdot \left|-\frac{a}{2} - (-a)\right|}{\left|\frac{2}{3} - \frac{3}{2}\right|} \]

4. **Simplify the Area Expression**:
- Calculate \( |c_1 - c_2| \):
\[ |c_1 - c_2| = \left|\frac{a}{3} - a\right| = \left|-\frac{2a}{3}\right| = \frac{2a}{3} \]
- Calculate \( |d_1 - d_2| \):
\[ |d_1 - d_2| = \left|-\frac{a}{2} + a\right| = \left| \frac{a}{2} \right| = \frac{a}{2} \]
- Calculate \( |m_1 - m_2| \):
\[ |m_1 - m_2| = \left|\frac{2}{3} - \frac{3}{2}\right| = \left| \frac{4 - 9}{6} \right| = \frac{5}{6} \]

5. **Substitute Back into Area Formula**:
\[ A = \frac{\left(\frac{2a}{3}\right) \cdot \left(\frac{a}{2}\right)}{\frac{5}{6}} = \frac{2a^2}{6 \cdot 5} = \frac{2a^2}{30} = \frac{a^2}{15} \]

6. **Set the Area Equal to 10**:
Since the area is given as 10 square units:
\[ \frac{a^2}{15} = 10 \]

7. **Solve for \( a \)**:
\[ a^2 = 150 \implies a = \pm \sqrt{150} = \pm 5\sqrt{6} \]

### Final Answer:
Thus, the value of \( a \) is \( \pm 5\sqrt{6} \).