Find the equations of the bisectors of the angles formed by the following pairs of lines
`x+2y+3=0` and `2x+y-2=0`

Answer

To find the equations of the angle bisectors formed by the lines \( x + 2y + 3 = 0 \) and \( 2x + y - 2 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients of the lines
The equations of the lines can be rewritten in the standard form \( Ax + By + C = 0 \):
- For the first line \( x + 2y + 3 = 0 \):
- \( A_1 = 1 \), \( B_1 = 2 \), \( C_1 = 3 \)

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

### Step 2: Use the angle bisector formula
The equations of the angle bisectors can be found using the formula:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]

### Step 3: Calculate the magnitudes of the coefficients
Calculate \( \sqrt{A_1^2 + B_1^2} \) and \( \sqrt{A_2^2 + B_2^2} \):
- For the first line:
\[ \sqrt{A_1^2 + B_1^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

- For the second line:
\[ \sqrt{A_2^2 + B_2^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 4: Substitute into the angle bisector formula
Substituting into the formula gives:
\[ \frac{1x + 2y + 3}{\sqrt{5}} = \pm \frac{2x + 1y - 2}{\sqrt{5}} \]

### Step 5: Remove the denominators
Multiplying through by \( \sqrt{5} \) to eliminate the denominators:
\[ 1x + 2y + 3 = \pm (2x + 1y - 2) \]

### Step 6: Solve for both cases (positive and negative)
**Case 1: Positive sign**
\[ x + 2y + 3 = 2x + y - 2 \]
Rearranging gives:
\[ x - 2y + 5 = 0 \quad \text{or} \quad x - 2y = -5 \]

**Case 2: Negative sign**
\[ x + 2y + 3 = - (2x + y - 2) \]
Rearranging gives:
\[ 3x + y + 1 = 0 \quad \text{or} \quad 3x + y = -1 \]

### Final Equations of the Angle Bisectors
The equations of the angle bisectors are:
1. \( x - 2y + 5 = 0 \)
2. \( 3x + y + 1 = 0 \)