Find the equations of the bisectors of the angles formed by the following pairs of lines
`3x+4y+13=0` and `12x-5y+32=0`

Answer

To find the equations of the angle bisectors formed by the lines \(3x + 4y + 13 = 0\) and \(12x - 5y + 32 = 0\), we will use the formula for the angle bisectors of two lines given by:

\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

### Step 1: Identify coefficients
From the given equations:
- For the first line \(3x + 4y + 13 = 0\):
- \(a_1 = 3\), \(b_1 = 4\), \(c_1 = 13\)

- For the second line \(12x - 5y + 32 = 0\):
- \(a_2 = 12\), \(b_2 = -5\), \(c_2 = 32\)

### Step 2: Calculate the denominators
Calculate \(\sqrt{a_1^2 + b_1^2}\) and \(\sqrt{a_2^2 + b_2^2}\):
- \(\sqrt{a_1^2 + b_1^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
- \(\sqrt{a_2^2 + b_2^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)

### Step 3: Set up the angle bisector equations
Using the formula:
\[ \frac{3x + 4y + 13}{5} = \pm \frac{12x - 5y + 32}{13} \]

### Step 4: Cross-multiply for both cases
1. **For the positive case:**
\[ 13(3x + 4y + 13) = 5(12x - 5y + 32) \]
Expanding both sides:
\[ 39x + 52y + 169 = 60x - 25y + 160 \]
Rearranging gives:
\[ 39x - 60x + 52y + 25y + 169 - 160 = 0 \]
\[ -21x + 77y + 9 = 0 \quad \text{(or)} \quad 21x - 77y - 9 = 0 \]

2. **For the negative case:**
\[ 13(3x + 4y + 13) = -5(12x - 5y + 32) \]
Expanding both sides:
\[ 39x + 52y + 169 = -60x + 25y - 160 \]
Rearranging gives:
\[ 39x + 60x + 52y - 25y + 169 + 160 = 0 \]
\[ 99x + 27y + 329 = 0 \]

### Step 5: Final equations of the angle bisectors
Thus, the equations of the angle bisectors are:
1. \(21x - 77y - 9 = 0\)
2. \(99x + 27y + 329 = 0\)