The area of the figure formed by the lines `ax + by +c = 0, ax - by + c = 0, ax+ by-c = 0` and `ax- by - c = 0` is

Options

A. `(c^(2))/(ab)`
B. `(2c^(2))/(ab)`
C. `(c^(2))/(2ab)`
D. `(c^(2))/(4ab)`

Explanation

To find the area of the figure formed by the lines \( ax + by + c = 0 \), \( ax - by + c = 0 \), \( ax + by - c = 0 \), and \( ax - by - c = 0 \), we can follow these steps:

### Step 1: Find the intersection points of the lines
1. **Line 1**: \( ax + by + c = 0 \)
- When \( x = 0 \), \( y = -\frac{c}{b} \) (Point A: \( (0, -\frac{c}{b}) \))
- When \( y = 0 \), \( x = -\frac{c}{a} \) (Point B: \( (-\frac{c}{a}, 0) \))

2. **Line 2**: \( ax - by + c = 0 \)
- When \( x = 0 \), \( y = \frac{c}{b} \) (Point C: \( (0, \frac{c}{b}) \))
- When \( y = 0 \), \( x = -\frac{c}{a} \) (Point D: \( (-\frac{c}{a}, 0) \))

3. **Line 3**: \( ax + by - c = 0 \)
- When \( x = 0 \), \( y = \frac{c}{b} \) (Point E: \( (0, \frac{c}{b}) \))
- When \( y = 0 \), \( x = \frac{c}{a} \) (Point F: \( (\frac{c}{a}, 0) \))

4. **Line 4**: \( ax - by - c = 0 \)
- When \( x = 0 \), \( y = -\frac{c}{b} \) (Point G: \( (0, -\frac{c}{b}) \))
- When \( y = 0 \), \( x = \frac{c}{a} \) (Point H: \( (\frac{c}{a}, 0) \))

### Step 2: Identify the vertices of the figure
The vertices of the figure formed by these lines are:
- \( A(0, -\frac{c}{b}) \)
- \( B(-\frac{c}{a}, 0) \)
- \( C(0, \frac{c}{b}) \)
- \( D(\frac{c}{a}, 0) \)

### Step 3: Calculate the area of the quadrilateral
The area of the quadrilateral can be calculated by dividing it into two triangles, \( \triangle ABC \) and \( \triangle ACD \).

#### Area of Triangle ABC:
Using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Base \( AC = \frac{c}{b} - (-\frac{c}{b}) = \frac{2c}{b} \)
- Height from point \( B \) to line \( AC \) is \( \frac{c}{a} \).

Thus, the area of triangle \( ABC \) is:
\[ \text{Area}_{ABC} = \frac{1}{2} \times \frac{2c}{b} \times \frac{c}{a} = \frac{c^2}{ab} \]

#### Area of Triangle ACD:
Similarly, the area of triangle \( ACD \) is the same as \( ABC \):
\[ \text{Area}_{ACD} = \frac{1}{2} \times \frac{2c}{b} \times \frac{c}{a} = \frac{c^2}{ab} \]

### Step 4: Total Area of the Figure
The total area of the figure formed by the lines is:
\[ \text{Total Area} = \text{Area}_{ABC} + \text{Area}_{ACD} = \frac{c^2}{ab} + \frac{c^2}{ab} = \frac{2c^2}{ab} \]

### Final Answer
Thus, the area of the figure formed by the lines is:
\[ \boxed{\frac{2c^2}{ab}} \]