The equation of the bisectors of the angles between the lines represented by the equation
`2(x + 2)^(2) + 3(x + 2)(y - 2) - 2(y - 2)^(2) = 0` is

Options

A. `3x^(2) - 8xy - 3y^(2) - 28 x + 4y + 32 = 0`
B. `3x^(2) + 8xy - 3y^(2) + 28 x - 4y + 32 = 0`
C. `3x^(2) - 8xy - 3y^(2) + 28 x - 4y + 32 = 0`
D. `3x^(2) - 8xy - 3y^(2) + 28 x - 4y - 32 = 0`

Answer

To solve the given problem, we need to find the equation of the angle bisectors of the lines represented by the equation:

\[ 2(x + 2)^2 + 3(x + 2)(y - 2) - 2(y - 2)^2 = 0 \]

### Step 1: Substitute Variables
Let \( a = x + 2 \) and \( b = y - 2 \). The equation can be rewritten as:

\[ 2a^2 + 3ab - 2b^2 = 0 \]

### Step 2: Factor the Quadratic Equation
We can factor the equation \( 2a^2 + 3ab - 2b^2 = 0 \).

Rearranging gives:

\[ 2a^2 + 4ab - ab - 2b^2 = 0 \]

Grouping terms:

\[ 2a(a + 2b) - b(a + 2b) = 0 \]

Factoring out \( (a + 2b) \):

\[ (a + 2b)(2a - b) = 0 \]

### Step 3: Set Each Factor to Zero
From the factored form, we have two equations:

1. \( a + 2b = 0 \)
2. \( 2a - b = 0 \)

### Step 4: Substitute Back for \( x \) and \( y \)
Now we substitute back \( a = x + 2 \) and \( b = y - 2 \).

#### For the first equation:
\[ a + 2b = 0 \]
\[ (x + 2) + 2(y - 2) = 0 \]
\[ x + 2 + 2y - 4 = 0 \]
\[ x + 2y - 2 = 0 \] (This is line \( l_1 \))

#### For the second equation:
\[ 2a - b = 0 \]
\[ 2(x + 2) - (y - 2) = 0 \]
\[ 2x + 4 - y + 2 = 0 \]
\[ 2x - y + 6 = 0 \] (This is line \( l_2 \))

### Step 5: Find the Angle Bisectors
The angle bisectors can be found using the formula:

\[ \frac{l_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{l_2}{\sqrt{A_2^2 + B_2^2}} \]

Where \( l_1 \) and \( l_2 \) are the equations of the lines, and \( A_1, B_1 \) and \( A_2, B_2 \) are the coefficients of \( x \) and \( y \) in those equations.

For line \( l_1: x + 2y - 2 = 0 \):
- \( A_1 = 1, B_1 = 2, C_1 = -2 \)

For line \( l_2: 2x - y + 6 = 0 \):
- \( A_2 = 2, B_2 = -1, C_2 = 6 \)

### Step 6: Apply the Formula
The equations of the angle bisectors become:

\[ \frac{x + 2y - 2}{\sqrt{1^2 + 2^2}} = \pm \frac{2x - y + 6}{\sqrt{2^2 + (-1)^2}} \]

Calculating the denominators:

\[ \sqrt{1 + 4} = \sqrt{5}, \quad \sqrt{4 + 1} = \sqrt{5} \]

Thus, we have:

\[ \frac{x + 2y - 2}{\sqrt{5}} = \pm \frac{2x - y + 6}{\sqrt{5}} \]

### Step 7: Remove the Denominator
Multiplying through by \( \sqrt{5} \):

\[ x + 2y - 2 = \pm (2x - y + 6) \]

### Step 8: Solve for Each Case
1. **For the positive case**:
\[ x + 2y - 2 = 2x - y + 6 \]
Rearranging gives:
\[ -x + 3y - 8 = 0 \quad \Rightarrow \quad x - 3y + 8 = 0 \]

2. **For the negative case**:
\[ x + 2y - 2 = - (2x - y + 6) \]
Rearranging gives:
\[ 3x + y - 4 = 0 \]

### Final Step: Combine the Results
The equations of the angle bisectors are:

1. \( x - 3y + 8 = 0 \)
2. \( 3x + y - 4 = 0 \)

### Conclusion
The final answer can be represented as the product of the two lines:

\[ (x - 3y + 8)(3x + y - 4) = 0 \]