Area of quadrilateral formed by striahgt lines `x =1, x = 3, y =2 and y = x + 3` is

Options

A. 6 sq. unit
B. 12 sq. unit
C. 3 sq. unit
D. None of these

Answer

To find the area of the quadrilateral formed by the lines \( x = 1 \), \( x = 3 \), \( y = 2 \), and \( y = x + 3 \), we will follow these steps:

### Step 1: Identify the Lines
The lines given are:
1. \( x = 1 \) (a vertical line)
2. \( x = 3 \) (another vertical line)
3. \( y = 2 \) (a horizontal line)
4. \( y = x + 3 \) (a diagonal line)

### Step 2: Find Intersection Points
We need to find the intersection points of these lines to determine the vertices of the quadrilateral.

- **Intersection of \( y = x + 3 \) and \( x = 1 \)**:
\[ y = 1 + 3 = 4 \quad \text{(Point A: (1, 4))} \]

- **Intersection of \( y = x + 3 \) and \( x = 3 \)**:
\[ y = 3 + 3 = 6 \quad \text{(Point B: (3, 6))} \]

- **Intersection of \( y = 2 \) and \( x = 1 \)**:
\[ \text{(Point C: (1, 2))} \]

- **Intersection of \( y = 2 \) and \( x = 3 \)**:
\[ \text{(Point D: (3, 2))} \]

### Step 3: Plot the Points
The vertices of the quadrilateral are:
- A (1, 4)
- B (3, 6)
- C (1, 2)
- D (3, 2)

### Step 4: Calculate the Area
The quadrilateral can be divided into a rectangle (ABCD) and a triangle (DCE).

1. **Area of Rectangle ABCD**:
- Length (AB) = \( 3 - 1 = 2 \)
- Height (from C to D) = \( 4 - 2 = 2 \)
\[ \text{Area of Rectangle} = \text{Length} \times \text{Height} = 2 \times 2 = 4 \text{ square units} \]

2. **Area of Triangle DCE**:
- Base (DC) = \( 3 - 1 = 2 \)
- Height (from E to D) = \( 6 - 4 = 2 \)
\[ \text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 2 = 2 \text{ square units} \]

### Step 5: Total Area of Quadrilateral
\[ \text{Total Area} = \text{Area of Rectangle} + \text{Area of Triangle} = 4 + 2 = 6 \text{ square units} \]

### Final Answer
The area of the quadrilateral formed by the lines is **6 square units**.

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