Let's break down the problem step by step:
1. **Finding the equations of lines \( L1 \) and \( L2 \)**:
- Line \( L1 \) passes through the point \((-2, -4)\) and intersects the x-axis at point \( A \) and the y-axis at point \( B \). Since it has a positive slope, we can write its equation in slope-intercept form: \( y = mx + b \).
- We know that \( B \) is the y-intercept, so \( B = (0, b) \).
- We need to find the slope \( m \) of \( L1 \) using the given point \((-2, -4)\).
- Once we have the equation of \( L1 \), we can find the equations of \( L2 \) since it's parallel to \( L1 \) and passes through \((0, -13)\).
2. **Finding the coordinates of \( A \) and \( B \)**:
- We can find the coordinates of \( A \) and \( B \) by substituting \( y = 0 \) to find \( A \) and \( x = 0 \) to find \( B \) in the equation of \( L1 \).
3. **Finding point \( C \)**:
- Since \( L2 \) is parallel to \( L1 \), it has the same slope.
- \( L2 \) passes through \((0, -13)\), so we can find its equation.
- We can find \( C \) by using the fact that \( CA = CB \), which means \( C \) lies on the perpendicular bisector of \( AB \).
4. **Calculating the area of triangle \( ABC \)**:
- Once we have the coordinates of \( A \), \( B \), and \( C \), we can use the formula for the area of a triangle given the coordinates of its vertices.
Let's proceed with these steps to find the area of triangle \( ABC \). If you need assistance with any specific step or calculation, feel free to ask!
