Find the equations of the bisectors of the angles between the lines `12x +5y - 4 = 0 and 3x + 4y + 7 = 0 ` .Prove that bisectors are at right angles to each other .

Answer

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

### Step 1: Identify the coefficients
From the first line \(12x + 5y - 4 = 0\), we have:
- \(A_1 = 12\)
- \(B_1 = 5\)
- \(C_1 = -4\)

From the second line \(3x + 4y + 7 = 0\), we have:
- \(A_2 = 3\)
- \(B_2 = 4\)
- \(C_2 = 7\)

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

### Step 3: Calculate the square roots
Calculate the denominators:
- \(\sqrt{A_1^2 + B_1^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)
- \(\sqrt{A_2^2 + B_2^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

### Step 4: Substitute values into the formula
Substituting the values into the angle bisector formula gives:
\[ \frac{12x + 5y - 4}{13} = \pm \frac{3x + 4y + 7}{5} \]

### Step 5: Solve for the positive case
Taking the positive case:
\[ \frac{12x + 5y - 4}{13} = \frac{3x + 4y + 7}{5} \]
Cross-multiplying gives:
\[ 5(12x + 5y - 4) = 13(3x + 4y + 7) \]
Expanding both sides:
\[ 60x + 25y - 20 = 39x + 52y + 91 \]
Rearranging gives:
\[ 60x - 39x + 25y - 52y - 20 - 91 = 0 \]
This simplifies to:
\[ 21x - 27y - 111 = 0 \]
Thus, the first angle bisector is:
\[ 21x - 27y - 111 = 0 \]

### Step 6: Solve for the negative case
Now taking the negative case:
\[ \frac{12x + 5y - 4}{13} = -\frac{3x + 4y + 7}{5} \]
Cross-multiplying gives:
\[ 5(12x + 5y - 4) = -13(3x + 4y + 7) \]
Expanding both sides:
\[ 60x + 25y - 20 = -39x - 52y - 91 \]
Rearranging gives:
\[ 60x + 39x + 25y + 52y - 20 + 91 = 0 \]
This simplifies to:
\[ 99x + 77y + 71 = 0 \]
Thus, the second angle bisector is:
\[ 99x + 77y + 71 = 0 \]

### Step 7: Prove that the bisectors are perpendicular
To prove that the bisectors are at right angles, we need to find the slopes of both bisectors.

- For the first bisector \(21x - 27y - 111 = 0\):
\[ 27y = 21x - 111 \implies y = \frac{21}{27}x - \frac{111}{27} \]
The slope \(m_1 = \frac{21}{27} = \frac{7}{9}\).

- For the second bisector \(99x + 77y + 71 = 0\):
\[ 77y = -99x - 71 \implies y = -\frac{99}{77}x - \frac{71}{77} \]
The slope \(m_2 = -\frac{99}{77}\).

### Step 8: Check the product of slopes
To check if the lines are perpendicular:
\[ m_1 \cdot m_2 = \frac{7}{9} \cdot \left(-\frac{99}{77}\right) = -\frac{693}{693} = -1 \]
Since the product of the slopes is \(-1\), the bisectors are perpendicular.

### Final Answer
The equations of the angle bisectors are:
1. \(21x - 27y - 111 = 0\)
2. \(99x + 77y + 71 = 0\)