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

Answer

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

### Step 1: Identify the coefficients
The equations of the lines can be expressed in the standard form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).
- For the first line \(4x + 3y - 4 = 0\):
- \(a_1 = 4\), \(b_1 = 3\), \(c_1 = -4\)

- For the second line \(12x + 5y - 3 = 0\):
- \(a_2 = 12\), \(b_2 = 5\), \(c_2 = -3\)

### Step 2: Apply the angle bisector formula
The equations of the angle bisectors are given by:
\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

Calculating the denominators:
- \(\sqrt{a_1^2 + b_1^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
- \(\sqrt{a_2^2 + b_2^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)

Substituting into the formula:
\[ \frac{4x + 3y - 4}{5} = \pm \frac{12x + 5y - 3}{13} \]

### Step 3: Cross-multiply
Cross-multiplying gives us two cases to solve:
1. \(13(4x + 3y - 4) = 5(12x + 5y - 3)\)
2. \(13(4x + 3y - 4) = -5(12x + 5y - 3)\)

#### Case 1:
\[ 13(4x + 3y - 4) = 5(12x + 5y - 3) \]
Expanding both sides:
\[ 52x + 39y - 52 = 60x + 25y - 15 \]
Rearranging gives:
\[ 52x + 39y - 60x - 25y + 15 - 52 = 0 \implies -8x + 14y - 37 = 0 \implies 8x - 14y + 37 = 0 \]

#### Case 2:
\[ 13(4x + 3y - 4) = -5(12x + 5y - 3) \]
Expanding both sides:
\[ 52x + 39y - 52 = -60x - 25y + 15 \]
Rearranging gives:
\[ 52x + 39y + 60x + 25y - 15 - 52 = 0 \implies 112x + 64y - 67 = 0 \]

### Step 4: Final equations of the bisectors
The equations of the angle bisectors are:
1. \(8x - 14y + 37 = 0\)
2. \(112x + 64y - 67 = 0\)

### Step 5: Show that the bisectors are perpendicular
To show that these two lines are perpendicular, we need to find their slopes.

#### Finding the slopes:
1. For \(8x - 14y + 37 = 0\):
\[ 14y = 8x + 37 \implies y = \frac{8}{14}x + \frac{37}{14} \implies m_1 = \frac{8}{14} = \frac{4}{7} \]

2. For \(112x + 64y - 67 = 0\):
\[ 64y = -112x + 67 \implies y = -\frac{112}{64}x + \frac{67}{64} \implies m_2 = -\frac{112}{64} = -\frac{7}{4} \]

### Step 6: Check the product of slopes
To check if the lines are perpendicular, we calculate:
\[ m_1 \cdot m_2 = \frac{4}{7} \cdot -\frac{7}{4} = -1 \]
Since the product of the slopes is \(-1\), the lines are perpendicular.

### Summary of the solution:
The equations of the angle bisectors are:
1. \(8x - 14y + 37 = 0\)
2. \(112x + 64y - 67 = 0\)

These bisectors are at right angles to each other.