Find the equations of the lines bisecting the angles between the lines `4x-3y+12=0 and 12x+5y=20`.

Answer

To find the equations of the lines bisecting the angles between the lines \(4x - 3y + 12 = 0\) and \(12x + 5y - 20 = 0\), we can use the formula for the angle bisectors of two lines given in the form \(Ax + By + C = 0\) and \(Px + Qy + R = 0\).

### Step-by-Step Solution:

1. **Identify the coefficients**:
- For the first line \(4x - 3y + 12 = 0\):
- \(A = 4\), \(B = -3\), \(C = 12\)
- For the second line \(12x + 5y - 20 = 0\):
- \(P = 12\), \(Q = 5\), \(R = -20\)

2. **Use the angle bisector formula**:
The angle bisectors can be found using the formula:
\[ \frac{Ax + By + C}{\sqrt{A^2 + B^2}} = \pm \frac{Px + Qy + R}{\sqrt{P^2 + Q^2}} \]

3. **Calculate the denominators**:
- For the first line:
\[ \sqrt{A^2 + B^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
- For the second line:
\[ \sqrt{P^2 + Q^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

4. **Set up the equation**:
Substitute the values into the angle bisector formula:
\[ \frac{4x - 3y + 12}{5} = \pm \frac{12x + 5y - 20}{13} \]

5. **Consider the positive case**:
\[ \frac{4x - 3y + 12}{5} = \frac{12x + 5y - 20}{13} \]
Cross-multiply:
\[ 13(4x - 3y + 12) = 5(12x + 5y - 20) \]
Expanding both sides:
\[ 52x - 39y + 156 = 60x + 25y - 100 \]
Rearranging gives:
\[ 52x - 60x - 39y - 25y + 156 + 100 = 0 \]
Simplifying:
\[ -8x - 64y + 256 = 0 \]
Dividing through by -8:
\[ x + 8y - 32 = 0 \]

6. **Consider the negative case**:
\[ \frac{4x - 3y + 12}{5} = -\frac{12x + 5y - 20}{13} \]
Cross-multiply:
\[ 13(4x - 3y + 12) = -5(12x + 5y - 20) \]
Expanding both sides:
\[ 52x - 39y + 156 = -60x - 25y + 100 \]
Rearranging gives:
\[ 52x + 60x - 39y + 25y + 156 - 100 = 0 \]
Simplifying:
\[ 112x - 14y + 56 = 0 \]
Dividing through by 14:
\[ 8x - y + 4 = 0 \]

### Final Equations of the Angle Bisectors:
1. \(x + 8y - 32 = 0\)
2. \(8x - y + 4 = 0\)