To solve the problem, we will follow these steps:
### Step 1: Find the intersection point of the angle bisectors
We have the equations of the angle bisectors:
1. \( x + y = 2 \) (Equation 1)
2. \( x - 3y = 6 \) (Equation 2)
We will solve these two equations simultaneously to find the coordinates of point \( I \).
**Solution:**
From Equation 1, we can express \( x \) in terms of \( y \):
\[
x = 2 - y
\]
Now, substitute \( x \) in Equation 2:
\[
(2 - y) - 3y = 6
\]
\[
2 - 4y = 6
\]
\[
-4y = 6 - 2
\]
\[
-4y = 4 \implies y = -1
\]
Now substitute \( y = -1 \) back into Equation 1 to find \( x \):
\[
x + (-1) = 2 \implies x = 3
\]
Thus, the coordinates of point \( I \) are \( (3, -1) \).
### Step 2: Find the image of point \( A \) across the bisectors
The coordinates of point \( A \) are \( (2, -4) \).
**Image of \( A \) across the line \( x - 3y = 6 \):**
We will use the formula for the reflection of a point across a line. The line can be written in the form \( Ax + By + C = 0 \):
\[
x - 3y - 6 = 0 \quad \text{(where } A = 1, B = -3, C = -6\text{)}
\]
Using the reflection formula:
\[
x' = x - \frac{2A(Ax + By + C)}{A^2 + B^2}
\]
\[
y' = y - \frac{2B(Ax + By + C)}{A^2 + B^2}
\]
Substituting \( (x, y) = (2, -4) \):
\[
Ax + By + C = 2 - 3(-4) - 6 = 2 + 12 - 6 = 8
\]
Now substituting into the reflection formulas:
\[
x' = 2 - \frac{2 \cdot 1 \cdot 8}{1^2 + (-3)^2} = 2 - \frac{16}{10} = 2 - 1.6 = 0.4
\]
\[
y' = -4 - \frac{2 \cdot (-3) \cdot 8}{1^2 + (-3)^2} = -4 + \frac{48}{10} = -4 + 4.8 = 0.8
\]
Thus, the image of point \( A \) across the line \( x - 3y = 6 \) is \( (0.4, 0.8) \).
### Step 3: Find the image of point \( A \) across the line \( x + y = 2 \)
Using the same reflection formula for the line \( x + y - 2 = 0 \):
\[
Ax + By + C = 2 - 4 - 2 = -4
\]
Now substituting into the reflection formulas:
\[
x' = 2 - \frac{2 \cdot 1 \cdot (-4)}{1^2 + 1^2} = 2 + \frac{8}{2} = 2 + 4 = 6
\]
\[
y' = -4 - \frac{2 \cdot 1 \cdot (-4)}{1^2 + 1^2} = -4 + \frac{8}{2} = -4 + 4 = 0
\]
Thus, the image of point \( A \) across the line \( x + y = 2 \) is \( (6, 0) \).
### Step 4: Find the coordinates of points \( B \) and \( C \)
From the previous steps, we have:
- Coordinates of \( B \) (image across \( x - 3y = 6 \)): \( (0.4, 0.8) \)
- Coordinates of \( C \) (image across \( x + y = 2 \)): \( (6, 0) \)
### Step 5: Calculate \( x_1x_2 + y_1y_2 \)
Let \( (x_1, y_1) = (0.4, 0.8) \) and \( (x_2, y_2) = (6, 0) \).
**Solution:**
\[
x_1x_2 + y_1y_2 = (0.4)(6) + (0.8)(0) = 2.4 + 0 = 2.4
\]
### Final Answer:
The value of \( x_1x_2 + y_1y_2 \) is \( 2.4 \).
---
