Find the equations of the bisectors of the angle between the straight lines 3x−4y+7=0 and 12x−5y−8=0

Options

A. a. 21x+27y−131=0
B. b. 99x−77y+51=0
C. c. x+8y +4=0
D. d. both (a) and (b)

Answer

To find the equations of the bisectors of the angle between the straight lines given by the equations \(3x - 4y + 7 = 0\) and \(12x - 5y - 8 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The equations of the lines can be expressed in the standard form \(Ax + By + C = 0\):
- 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:
- \(A_2 = 12\), \(B_2 = -5\), \(C_2 = -8\)

### Step 2: Use the angle bisector formula
The equations of the angle bisectors between two lines can be given by the formula:
\[ \frac{A_1}{\sqrt{A_1^2 + B_1^2}} \cdot \frac{A_2}{\sqrt{A_2^2 + B_2^2}} + \frac{B_1}{\sqrt{A_1^2 + B_1^2}} \cdot \frac{B_2}{\sqrt{A_2^2 + B_2^2}} = \pm \frac{C_1}{\sqrt{A_1^2 + B_1^2}} \cdot \frac{C_2}{\sqrt{A_2^2 + B_2^2}} \]

### Step 3: Calculate the necessary values
First, we 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:
\[ \sqrt{A_2^2 + B_2^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

### Step 4: Substitute into the angle bisector formula
Now, substituting into the formula:
\[ \frac{3}{5} \cdot \frac{12}{13} + \frac{-4}{5} \cdot \frac{-5}{13} = \pm \frac{7}{5} \cdot \frac{-8}{13} \]

Calculating the left side:
\[ \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \]

Calculating the right side:
\[ \pm \frac{-56}{65} \]

### Step 5: Form the equations
This gives us two equations:
1. \(56 = -56\) (not valid)
2. \(56 = 56\)

Thus, we can write the equations of the bisectors as:
\[ \frac{3x - 4y + 7}{5} + \frac{12x - 5y - 8}{13} = 0 \]
and
\[ \frac{3x - 4y + 7}{5} - \frac{12x - 5y - 8}{13} = 0 \]

### Step 6: Simplify the equations
1. For the first equation:
\[ 13(3x - 4y + 7) + 5(12x - 5y - 8) = 0 \]
Simplifying gives:
\[ 39x - 52y + 91 + 60x - 25y - 40 = 0 \implies 99x - 77y + 51 = 0 \]

2. For the second equation:
\[ 13(3x - 4y + 7) - 5(12x - 5y - 8) = 0 \]
Simplifying gives:
\[ 39x - 52y + 91 - 60x + 25y + 40 = 0 \implies -21x - 27y + 131 = 0 \]

### Final Result
The equations of the angle bisectors are:
1. \(99x - 77y + 51 = 0\)
2. \(-21x - 27y + 131 = 0\)