To find the bisector of the exterior angle at point C of the triangle formed by the lines given by the equations \( AB: x + y - 5 = 0 \), \( BC: x + 7y - 7 = 0 \), and \( CA: 7x + y + 14 = 0 \), we can follow these steps:
### Step 1: Identify the equations of the lines
We have the following equations:
1. \( AB: x + y - 5 = 0 \)
2. \( BC: x + 7y - 7 = 0 \)
3. \( CA: 7x + y + 14 = 0 \)
### Step 2: Find the points of intersection of the lines
To find the vertices of the triangle, we need to calculate the intersection points of these lines.
**Finding point A (intersection of AB and CA):**
We solve the equations:
- \( x + y - 5 = 0 \) (1)
- \( 7x + y + 14 = 0 \) (2)
From (1), we can express \( y \) in terms of \( x \):
\[ y = 5 - x \]
Substituting into (2):
\[ 7x + (5 - x) + 14 = 0 \]
\[ 7x - x + 19 = 0 \]
\[ 6x + 19 = 0 \]
\[ x = -\frac{19}{6} \]
Now substituting \( x \) back into (1) to find \( y \):
\[ y = 5 - \left(-\frac{19}{6}\right) = \frac{30}{6} + \frac{19}{6} = \frac{49}{6} \]
Thus, point A is \( \left(-\frac{19}{6}, \frac{49}{6}\right) \).
**Finding point B (intersection of AB and BC):**
We solve:
- \( x + y - 5 = 0 \) (1)
- \( x + 7y - 7 = 0 \) (3)
From (1):
\[ y = 5 - x \]
Substituting into (3):
\[ x + 7(5 - x) - 7 = 0 \]
\[ x + 35 - 7x - 7 = 0 \]
\[ -6x + 28 = 0 \]
\[ x = \frac{28}{6} = \frac{14}{3} \]
Now substituting \( x \) back into (1) to find \( y \):
\[ y = 5 - \frac{14}{3} = \frac{15}{3} - \frac{14}{3} = \frac{1}{3} \]
Thus, point B is \( \left(\frac{14}{3}, \frac{1}{3}\right) \).
**Finding point C (intersection of BC and CA):**
We solve:
- \( x + 7y - 7 = 0 \) (3)
- \( 7x + y + 14 = 0 \) (2)
From (3):
\[ y = \frac{7 - x}{7} \]
Substituting into (2):
\[ 7x + \frac{7 - x}{7} + 14 = 0 \]
Multiplying through by 7 to eliminate the fraction:
\[ 49x + 7 - x + 98 = 0 \]
\[ 48x + 105 = 0 \]
\[ x = -\frac{105}{48} \]
Now substituting \( x \) back into (3) to find \( y \):
\[ y = \frac{7 - \left(-\frac{105}{48}\right)}{7} = \frac{7 + \frac{105}{48}}{7} = \frac{\frac{336 + 105}{48}}{7} = \frac{441}{336} = \frac{21}{16} \]
Thus, point C is \( \left(-\frac{105}{48}, \frac{21}{16}\right) \).
### Step 3: Find the angle bisector of the exterior angle at C
The angle bisector of the exterior angle at point C can be found using the formula for the angle bisector between two lines. The equations of the lines \( AC \) and \( BC \) are:
- \( AC: 7x + y + 14 = 0 \)
- \( BC: x + 7y - 7 = 0 \)
Using the angle bisector formula:
\[
\frac{Ax + By + C}{\sqrt{A^2 + B^2}} = \pm \frac{A'x + B'y + C'}{\sqrt{A'^2 + B'^2}}
\]
Where \( A, B, C \) are the coefficients from the first line and \( A', B', C' \) from the second line.
Substituting:
- For \( AC: A = 7, B = 1, C = 14 \)
- For \( BC: A' = 1, B' = 7, C' = -7 \)
Thus, we have:
\[
\frac{7x + y + 14}{\sqrt{7^2 + 1^2}} = \pm \frac{x + 7y - 7}{\sqrt{1^2 + 7^2}}
\]
Calculating the denominators:
\[
\sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}
\]
\[
\sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}
\]
Thus, we can simplify the equation:
\[
7x + y + 14 = \pm (x + 7y - 7)
\]
This gives us two equations to solve:
1. \( 7x + y + 14 = x + 7y - 7 \)
2. \( 7x + y + 14 = - (x + 7y - 7) \)
### Step 4: Solve the equations
**For the first equation:**
\[
7x + y + 14 = x + 7y - 7
\]
Rearranging gives:
\[
6x - 6y + 21 = 0 \quad \Rightarrow \quad 2x - 2y + 7 = 0 \quad \Rightarrow \quad 2x + 7 = 2y \quad \Rightarrow \quad y = x + \frac{7}{2}
\]
**For the second equation:**
\[
7x + y + 14 = -x - 7y + 7
\]
Rearranging gives:
\[
8x + 8y + 7 = 0 \quad \Rightarrow \quad x + y + \frac{7}{8} = 0 \quad \Rightarrow \quad y = -x - \frac{7}{8}
\]
### Step 5: Determine which bisector is external
To determine which of the two lines is the external bisector, we can check the position of the points A and B relative to each line.
Calculating for point A in both equations:
1. For \( 2x - 2y + 7 = 0 \):
- Substitute \( A \left(-\frac{19}{6}, \frac{49}{6}\right) \)
- If the result is positive, point A lies on one side of the line.
2. For \( 8x + 8y + 7 = 0 \):
- Substitute \( A \left(-\frac{19}{6}, \frac{49}{6}\right) \)
- If the result is positive, point A lies on the same side of the line.
After checking both lines with points A and B, we find that the line \( 8x + 8y + 7 = 0 \) is the external angle bisector.
### Final Answer:
The bisector of the exterior angle at C is:
\[
\boxed{8x + 8y + 7 = 0}
\]
