A triangle is formed by the lines whose equations are `AB:x+y-5=0, BC:x+7y-7=0, and CA:7x+y+14=0`. Find the bisector of the interior angle at B

Answer

To find the bisector of the interior angle at point B formed by the lines AB, BC, and CA, we can follow these steps:

### Step 1: Identify the equations of the lines
The equations of the lines forming the triangle are:
- Line AB: \( x + y - 5 = 0 \)
- Line BC: \( x + 7y - 7 = 0 \)
- Line CA: \( 7x + y + 14 = 0 \)

### Step 2: Find the slopes of the lines
To find the angle bisector at B, we first need to determine the slopes of lines AB and BC.

For line AB:
- The equation is in the form \( Ax + By + C = 0 \) where \( A = 1, B = 1, C = -5 \).
- The slope \( m_1 \) of line AB is given by \( m_1 = -\frac{A}{B} = -\frac{1}{1} = -1 \).

For line BC:
- The equation is \( x + 7y - 7 = 0 \) where \( A = 1, B = 7, C = -7 \).
- The slope \( m_2 \) of line BC is \( m_2 = -\frac{1}{7} \).

### Step 3: Find the angle between the lines
To check if angle B is acute or obtuse, we can use the formula for the tangent of the angle between two lines:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Substituting the slopes:
\[ \tan B = \left| \frac{-\frac{1}{7} - (-1)}{1 + (-1)(-\frac{1}{7})} \right| = \left| \frac{-\frac{1}{7} + 1}{1 + \frac{1}{7}} \right| = \left| \frac{\frac{6}{7}}{\frac{8}{7}} \right| = \frac{6}{8} = \frac{3}{4} \]
Since \( \tan B \) is positive, angle B is acute.

### Step 4: Use the angle bisector formula
The equation of the angle bisector of the lines given by:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]
For lines AB and BC:
- For line AB: \( A_1 = 1, B_1 = 1, C_1 = -5 \)
- For line BC: \( A_2 = 1, B_2 = 7, C_2 = -7 \)

### Step 5: Calculate the left-hand side and right-hand side
The left-hand side becomes:
\[ \frac{x + y - 5}{\sqrt{1^2 + 1^2}} = \frac{x + y - 5}{\sqrt{2}} \]
The right-hand side becomes:
\[ \frac{x + 7y - 7}{\sqrt{1^2 + 7^2}} = \frac{x + 7y - 7}{\sqrt{50}} = \frac{x + 7y - 7}{5\sqrt{2}} \]

### Step 6: Set up the equation
Setting the two sides equal (using the negative sign since \( A_1A_2 + B_1B_2 > 0 \)):
\[ \frac{x + y - 5}{\sqrt{2}} = -\frac{x + 7y - 7}{5\sqrt{2}} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 5(x + y - 5) = - (x + 7y - 7) \]
Expanding both sides:
\[ 5x + 5y - 25 = -x - 7y + 7 \]
Rearranging gives:
\[ 6x + 12y - 32 = 0 \]

### Step 8: Simplify the equation
Dividing the entire equation by 2:
\[ 3x + 6y - 16 = 0 \]

Thus, the equation of the bisector of the interior angle at B is:
\[ 3x + 6y - 16 = 0 \]