Area of quadrilateral formed by straight lines `2x =- 5 , 2y = 3 , x=1 and y + 2 = 0` is

Options

A. `(21)/(2)` sq. unit
B. `(49)/(4)` sq. unit
C. `(21)/(8)` sq. unit
D. `(7)/(4)` sq. unit

Answer

To find the area of the quadrilateral formed by the lines \(2x = -5\), \(2y = 3\), \(x = 1\), and \(y + 2 = 0\), we will follow these steps:

### Step 1: Convert the equations into slope-intercept form or identify their characteristics
1. **Equation 1**: \(2x = -5\) can be simplified to \(x = -\frac{5}{2}\). This line is vertical.
2. **Equation 2**: \(2y = 3\) simplifies to \(y = \frac{3}{2}\). This line is horizontal.
3. **Equation 3**: \(x = 1\) is another vertical line.
4. **Equation 4**: \(y + 2 = 0\) simplifies to \(y = -2\). This line is also horizontal.

### Step 2: Identify the intersection points of the lines
- **Point A**: Intersection of \(x = 1\) and \(y = \frac{3}{2}\) is \((1, \frac{3}{2})\).
- **Point B**: Intersection of \(x = 1\) and \(y = -2\) is \((1, -2)\).
- **Point C**: Intersection of \(x = -\frac{5}{2}\) and \(y = -2\) is \((- \frac{5}{2}, -2)\).
- **Point D**: Intersection of \(x = -\frac{5}{2}\) and \(y = \frac{3}{2}\) is \((- \frac{5}{2}, \frac{3}{2})\).

### Step 3: List the coordinates of the vertices
- \(A(1, \frac{3}{2})\)
- \(B(1, -2)\)
- \(C(-\frac{5}{2}, -2)\)
- \(D(-\frac{5}{2}, \frac{3}{2})\)

### Step 4: Calculate the lengths of the sides
- **Length AB**: The distance between points A and B can be calculated as:
\[ AB = |y_1 - y_2| = \left| \frac{3}{2} - (-2) \right| = \left| \frac{3}{2} + 2 \right| = \left| \frac{3}{2} + \frac{4}{2} \right| = \left| \frac{7}{2} \right| = \frac{7}{2} \]

- **Length AD**: The distance between points A and D can be calculated as:
\[ AD = |x_1 - x_2| = |1 - (-\frac{5}{2})| = |1 + \frac{5}{2}| = |1 + 2.5| = |3.5| = \frac{7}{2} \]

### Step 5: Calculate the area of the quadrilateral
Since both lengths are equal, the quadrilateral is a rectangle. The area \(A\) can be calculated as:
\[ \text{Area} = \text{Length} \times \text{Width} = AB \times AD = \frac{7}{2} \times \frac{7}{2} = \frac{49}{4} \text{ square units} \]

### Final Answer
The area of the quadrilateral is \(\frac{49}{4}\) square units.

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