The area defined by |2x + y| + |x - 2| `le` 4 in the x - y coordinate plane is

Options

A. 32
B. `(32)/(5)`
C. 16
D. none of these

Answer

To solve the problem of finding the area defined by the inequality \( |2x + y| + |x - 2| \leq 4 \) in the x-y coordinate plane, we will break it down into cases based on the expressions inside the absolute values.

### Step 1: Identify Cases
We need to consider the different cases based on the signs of the expressions \( 2x + y \) and \( x - 2 \).

1. **Case 1**: \( 2x + y \geq 0 \) and \( x - 2 \geq 0 \)
- Here, the inequality simplifies to:
\[ (2x + y) + (x - 2) \leq 4 \implies 3x + y \leq 6 \]

2. **Case 2**: \( 2x + y \geq 0 \) and \( x - 2 < 0 \)
- The inequality simplifies to:
\[ (2x + y) - (x - 2) \leq 4 \implies x + y \leq 2 \]

3. **Case 3**: \( 2x + y < 0 \) and \( x - 2 \geq 0 \)
- The inequality simplifies to:
\[ -(2x + y) + (x - 2) \leq 4 \implies -2x - y + x - 2 \leq 4 \implies -x - y \leq 6 \implies x + y \geq -6 \]

4. **Case 4**: \( 2x + y < 0 \) and \( x - 2 < 0 \)
- The inequality simplifies to:
\[ -(2x + y) - (x - 2) \leq 4 \implies -2x - y - x + 2 \leq 4 \implies -3x - y \leq 2 \implies 3x + y \geq -2 \]

### Step 2: Write the Equations
From the above cases, we have the following linear equations:

1. \( 3x + y = 6 \) (from Case 1)
2. \( x + y = 2 \) (from Case 2)
3. \( x + y = -6 \) (from Case 3)
4. \( 3x + y = -2 \) (from Case 4)

### Step 3: Find Intersection Points
Next, we will find the intersection points of these lines to determine the vertices of the region defined by the inequalities.

1. **Intersection of \( 3x + y = 6 \) and \( x + y = 2 \)**:
- Solve:
\[ y = 2 - x \quad \text{(from the second equation)} \]
Substitute into the first:
\[ 3x + (2 - x) = 6 \implies 2x = 4 \implies x = 2, \quad y = 0 \quad \text{(Point: (2, 0))} \]

2. **Intersection of \( 3x + y = 6 \) and \( x + y = -6 \)**:
- Solve:
\[ y = -6 - x \quad \text{(from the second equation)} \]
Substitute into the first:
\[ 3x + (-6 - x) = 6 \implies 2x = 12 \implies x = 6, \quad y = -12 \quad \text{(Point: (6, -12))} \]

3. **Intersection of \( x + y = 2 \) and \( 3x + y = -2 \)**:
- Solve:
\[ y = 2 - x \quad \text{(from the first equation)} \]
Substitute into the second:
\[ 3x + (2 - x) = -2 \implies 2x = -4 \implies x = -2, \quad y = 4 \quad \text{(Point: (-2, 4))} \]

4. **Intersection of \( x + y = -6 \) and \( 3x + y = -2 \)**:
- Solve:
\[ y = -6 - x \quad \text{(from the first equation)} \]
Substitute into the second:
\[ 3x + (-6 - x) = -2 \implies 2x = 4 \implies x = 2, \quad y = -8 \quad \text{(Point: (2, -8))} \]

### Step 4: Determine the Area
Now we have the vertices of the region defined by the inequalities. The vertices are:
- (2, 0)
- (6, -12)
- (-2, 4)
- (2, -8)

To find the area of the polygon formed by these vertices, we can use the formula for the area of a polygon based on its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \]

Substituting the coordinates into the formula gives us the area.

### Final Calculation
After performing the calculations, we find that the area is \( 32 \) square units.