To find the area enclosed by the lines \( y = 0 \), \( x + 6 = 0 \), and \( 2x - 3y = 6 \), we will follow these steps:
### Step 1: Identify the equations of the lines
1. The first line is \( y = 0 \) (the x-axis).
2. The second line is \( x + 6 = 0 \), which simplifies to \( x = -6 \) (a vertical line).
3. The third line is \( 2x - 3y = 6 \). We can rearrange this to find \( y \) in terms of \( x \):
\[
3y = 2x - 6 \implies y = \frac{2}{3}x - 2
\]
### Step 2: Find the points of intersection
Next, we will find the points where these lines intersect.
1. **Intersection of \( y = 0 \) and \( x = -6 \)**:
- Substitute \( y = 0 \) into \( x = -6 \):
- The point is \( (-6, 0) \).
2. **Intersection of \( y = 0 \) and \( 2x - 3y = 6 \)**:
- Substitute \( y = 0 \) into \( 2x - 3(0) = 6 \):
- \( 2x = 6 \implies x = 3 \).
- The point is \( (3, 0) \).
3. **Intersection of \( x = -6 \) and \( 2x - 3y = 6 \)**:
- Substitute \( x = -6 \) into \( 2(-6) - 3y = 6 \):
- \( -12 - 3y = 6 \implies -3y = 18 \implies y = -6 \).
- The point is \( (-6, -6) \).
### Step 3: Identify the vertices of the enclosed area
The vertices of the triangle formed by these intersections are:
1. \( (-6, 0) \)
2. \( (3, 0) \)
3. \( (-6, -6) \)
### Step 4: Calculate the area of the triangle
To find the area of the triangle formed by these vertices, we can use the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is the distance between the points \( (-6, 0) \) and \( (3, 0) \), and the height is the vertical distance from the point \( (-6, -6) \) to the x-axis.
1. **Base**: The distance between \( (-6, 0) \) and \( (3, 0) \) is:
\[
|3 - (-6)| = 3 + 6 = 9
\]
2. **Height**: The vertical distance from \( (-6, -6) \) to the x-axis (y = 0) is:
\[
|-6 - 0| = 6
\]
Now, substituting these values into the area formula:
\[
\text{Area} = \frac{1}{2} \times 9 \times 6 = \frac{54}{2} = 27
\]
### Final Answer
The area enclosed by the lines is \( \boxed{27} \) square units.
