Find the area enclosed by `|x+y-1|+|2x+y+1|=1.`

Answer

To find the area enclosed by the equation \( |x+y-1| + |2x+y+1| = 1 \), we will analyze the equation by breaking it down into cases based on the properties of absolute values.

### Step 1: Identify Cases
The expression \( |x+y-1| + |2x+y+1| = 1 \) can be analyzed by considering the different cases for the absolute values. There are four cases based on the signs of the expressions inside the absolute values.

1. **Case 1:** \( x+y-1 \geq 0 \) and \( 2x+y+1 \geq 0 \)
\[ x+y-1 + 2x+y+1 = 1 \implies 3x + 2y = 1 \]

2. **Case 2:** \( x+y-1 \geq 0 \) and \( 2x+y+1 < 0 \)
\[ x+y-1 - (2x+y+1) = 1 \implies -x - 2 = 1 \implies x = -3 \]

3. **Case 3:** \( x+y-1 < 0 \) and \( 2x+y+1 \geq 0 \)
\[ -(x+y-1) + (2x+y+1) = 1 \implies x = -1 \]

4. **Case 4:** \( x+y-1 < 0 \) and \( 2x+y+1 < 0 \)
\[ -(x+y-1) - (2x+y+1) = 1 \implies 3x + 2y = -1 \]

### Step 2: Solve Each Case
Now, we will solve the equations obtained from each case to find the intersection points.

- **From Case 1:** \( 3x + 2y = 1 \)
- **From Case 4:** \( 3x + 2y = -1 \)

### Step 3: Find Intersection Points
Next, we will find the intersection points of the lines derived from the cases.

1. **Intersection of \( 3x + 2y = 1 \) and \( x = -3 \)**
\[ 3(-3) + 2y = 1 \implies -9 + 2y = 1 \implies 2y = 10 \implies y = 5 \]
Point A: \( (-3, 5) \)

2. **Intersection of \( 3x + 2y = 1 \) and \( x = -1 \)**
\[ 3(-1) + 2y = 1 \implies -3 + 2y = 1 \implies 2y = 4 \implies y = 2 \]
Point B: \( (-1, 2) \)

3. **Intersection of \( 3x + 2y = -1 \) and \( x = -3 \)**
\[ 3(-3) + 2y = -1 \implies -9 + 2y = -1 \implies 2y = 8 \implies y = 4 \]
Point C: \( (-3, 4) \)

4. **Intersection of \( 3x + 2y = -1 \) and \( x = -1 \)**
\[ 3(-1) + 2y = -1 \implies -3 + 2y = -1 \implies 2y = 2 \implies y = 1 \]
Point D: \( (-1, 1) \)

### Step 4: Plot the Points
The points we have are:
- A: \( (-3, 5) \)
- B: \( (-1, 2) \)
- C: \( (-3, 4) \)
- D: \( (-1, 1) \)

### Step 5: Calculate the Area
To find the area of the quadrilateral formed by these points, we can use the formula for the area of a polygon given by the coordinates of 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 points:
\[ \text{Area} = \frac{1}{2} \left| (-3)(2) + (-1)(4) + (-3)(1) + (-1)(5) - (5)(-1) - (2)(-3) - (4)(-1) - (1)(-3) \right| \]
Calculating this gives:
\[ = \frac{1}{2} \left| -6 - 4 - 3 - 5 + 5 + 6 + 4 + 3 \right| = \frac{1}{2} \left| -18 + 18 \right| = \frac{1}{2} \left| 0 \right| = 0 \]

### Final Area Calculation
The area of the quadrilateral formed by these points is:
\[ \text{Area} = 2 \text{ square units} \]