The equations of the bisectors of the angles between the straight line `3x-4y+7=0` and `12x-5y-8=0`, are:

Options

A. `21x+27y+131=0`
B. `x+27y-131=0`
C. `21x-27+131=0`
D. `21x+27y-131=0`

Answer

To find the equations of the bisectors of the angles between the straight lines given by the equations \(3x - 4y + 7 = 0\) and \(12x - 5y - 8 = 0\), we can use the formula for the angle bisectors of two lines.

### Step-by-Step Solution:

1. **Identify the coefficients**:
For the first line \(3x - 4y + 7 = 0\), we have:
- \(A_1 = 3\)
- \(B_1 = -4\)
- \(C_1 = 7\)

For the second line \(12x - 5y - 8 = 0\), we have:
- \(A_2 = 12\)
- \(B_2 = -5\)
- \(C_2 = -8\)

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

3. **Calculate the denominators**:
- For the first line:
\[ \sqrt{A_1^2 + B_1^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- For the second line:
\[ \sqrt{A_2^2 + B_2^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

4. **Set up the equation**:
Plugging into the bisector formula gives:
\[ \frac{3x - 4y + 7}{5} = \pm \frac{12x - 5y - 8}{13} \]

5. **Cross-multiply**:
This results in two equations:
- Positive case:
\[ 13(3x - 4y + 7) = 5(12x - 5y - 8) \]
- Negative case:
\[ 13(3x - 4y + 7) = -5(12x - 5y - 8) \]

6. **Solve the positive case**:
Expanding the positive case:
\[ 39x - 52y + 91 = 60x - 25y - 40 \]
Rearranging gives:
\[ 60x - 39x - 25y + 52y - 40 - 91 = 0 \]
Simplifying:
\[ 21x + 27y - 131 = 0 \]

7. **Solve the negative case**:
Expanding the negative case:
\[ 39x - 52y + 91 = -60x + 25y + 40 \]
Rearranging gives:
\[ 39x + 60x - 52y - 25y + 91 - 40 = 0 \]
Simplifying:
\[ 99x - 77y + 51 = 0 \]

8. **Final equations of the bisectors**:
The equations of the bisectors are:
- \(21x + 27y - 131 = 0\) (first bisector)
- \(99x - 77y + 51 = 0\) (second bisector)