Find the equations of the bisectors of the angles formed by the following pairs of lines
`4x+3y-5=0` and `5x+12y-41=0`

Answer

To find the equations of the angle bisectors formed by the lines \(4x + 3y - 5 = 0\) and \(5x + 12y - 41 = 0\), we will use the formula for the angle bisectors of two lines given in the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).

### Step 1: Identify coefficients
For the given lines:
1. Line 1: \(4x + 3y - 5 = 0\)
- Here, \(a_1 = 4\), \(b_1 = 3\), \(c_1 = -5\)
2. Line 2: \(5x + 12y - 41 = 0\)
- Here, \(a_2 = 5\), \(b_2 = 12\), \(c_2 = -41\)

### Step 2: Apply the angle bisector formula
The formula for the angle bisectors is 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 3: Calculate the denominators
Calculate \(\sqrt{a_1^2 + b_1^2}\) and \(\sqrt{a_2^2 + b_2^2}\):
- For line 1:
\[ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
- For line 2:
\[ \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

### Step 4: Substitute into the formula
Substituting the values into the angle bisector formula:
\[ \frac{4x + 3y - 5}{5} = \pm \frac{5x + 12y - 41}{13} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 13(4x + 3y - 5) = \pm 5(5x + 12y - 41) \]

### Step 6: Solve for the positive case
First, consider the positive case:
\[ 13(4x + 3y - 5) = 5(5x + 12y - 41) \]
Expanding both sides:
\[ 52x + 39y - 65 = 25x + 60y - 205 \]
Rearranging gives:
\[ 52x - 25x + 39y - 60y = -205 + 65 \]
This simplifies to:
\[ 27x - 21y + 140 = 0 \]

### Step 7: Solve for the negative case
Now, consider the negative case:
\[ 13(4x + 3y - 5) = -5(5x + 12y - 41) \]
Expanding both sides:
\[ 52x + 39y - 65 = -25x - 60y + 205 \]
Rearranging gives:
\[ 52x + 25x + 39y + 60y = 205 + 65 \]
This simplifies to:
\[ 77x + 99y - 270 = 0 \]

### Final Result
The equations of the angle bisectors are:
1. \(27x - 21y + 140 = 0\)
2. \(77x + 99y - 270 = 0\)