The area bounded by the straight lines `y =1` and `+- 2 x + y = 2,` in square units, is

Options

A. `1//2`
B. 1
C. `3//2`
D. 2

Answer

To find the area bounded by the straight lines \(y = 1\) and \(2x + y = 2\), and \(-2x + y = 2\), we can follow these steps:

### Step 1: Identify the equations of the lines
We have three lines:
1. \(y = 1\) (Line 1)
2. \(2x + y = 2\) (Line 2)
3. \(-2x + y = 2\) (Line 3)

### Step 2: Find the intersection points of the lines
To find the area bounded by these lines, we need to determine their intersection points.

**Intersection of Line 1 and Line 2:**
Substituting \(y = 1\) into \(2x + y = 2\):
\[ 2x + 1 = 2 \implies 2x = 1 \implies x = \frac{1}{2} \]
Thus, the intersection point is \(\left(\frac{1}{2}, 1\right)\).

**Intersection of Line 1 and Line 3:**
Substituting \(y = 1\) into \(-2x + y = 2\):
\[ -2x + 1 = 2 \implies -2x = 1 \implies x = -\frac{1}{2} \]
Thus, the intersection point is \(\left(-\frac{1}{2}, 1\right)\).

**Intersection of Line 2 and Line 3:**
To find the intersection of \(2x + y = 2\) and \(-2x + y = 2\), we can set the equations equal to each other:
\[ 2x + y = 2 \quad \text{(1)} \]
\[ -2x + y = 2 \quad \text{(2)} \]
Subtracting (2) from (1):
\[ (2x + y) - (-2x + y) = 2 - 2 \implies 4x = 0 \implies x = 0 \]
Substituting \(x = 0\) into either equation (let's use (1)):
\[ 2(0) + y = 2 \implies y = 2 \]
Thus, the intersection point is \((0, 2)\).

### Step 3: Identify the vertices of the triangle
The vertices of the triangle formed by the intersection points are:
- Point A: \(\left(\frac{1}{2}, 1\right)\)
- Point B: \(\left(-\frac{1}{2}, 1\right)\)
- Point C: \((0, 2)\)

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points A, B, and C:
\[ A = \frac{1}{2} \left| \frac{1}{2}(1 - 2) + \left(-\frac{1}{2}\right)(2 - 1) + 0(1 - 1) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| \frac{1}{2}(-1) + \left(-\frac{1}{2}\right)(1) + 0 \right| \]
\[ = \frac{1}{2} \left| -\frac{1}{2} - \frac{1}{2} \right| = \frac{1}{2} \left| -1 \right| = \frac{1}{2} \times 1 = \frac{1}{2} \]

### Final Answer
The area bounded by the straight lines is \(\frac{1}{2}\) square units.

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