If the area of the rhombus enclosed by the lines `xpmypmn=0` be 2 square units, then

Options

A. `n^(2)=4`
B. `n^(2)=2`
C. `n^(2)=(1)/(2)`
D. `n^(2)=1`

Answer

To solve the problem, we need to find the condition for the area of the rhombus formed by the lines \( x + y = n \), \( x - y = n \), \( -x + y = n \), and \( -x - y = n \) to be equal to 2 square units.

### Step-by-Step Solution:

1. **Identify the equations of the lines**:
The equations given are:
\[ x + y = n, \quad x - y = n, \quad -x + y = n, \quad -x - y = n \]

2. **Find the vertices of the rhombus**:
To find the vertices, we can solve the pairs of equations:
- For \( x + y = n \) and \( x - y = n \):
\[ \begin{align*} x + y &= n \quad (1) \\ x - y &= n \quad (2) \end{align*} \]
Adding (1) and (2):
\[ 2x = 2n \implies x = n \]
Substituting \( x = n \) in (1):
\[ n + y = n \implies y = 0 \]
So one vertex is \( (n, 0) \).

- For \( x + y = n \) and \( -x + y = n \):
\[ \begin{align*} x + y &= n \quad (3) \\ -x + y &= n \quad (4) \end{align*} \]
Adding (3) and (4):
\[ 2y = 2n \implies y = n \]
Substituting \( y = n \) in (3):
\[ x + n = n \implies x = 0 \]
So another vertex is \( (0, n) \).

- Continuing this process, we find the other vertices:
- \( (-n, 0) \)
- \( (0, -n) \)

Thus, the vertices of the rhombus are \( (n, 0) \), \( (0, n) \), \( (-n, 0) \), and \( (0, -n) \).

3. **Calculate the lengths of the diagonals**:
The diagonals of the rhombus are:
- \( d_1 \) (horizontal diagonal) between \( (n, 0) \) and \( (-n, 0) \):
\[ d_1 = n - (-n) = 2n \]
- \( d_2 \) (vertical diagonal) between \( (0, n) \) and \( (0, -n) \):
\[ d_2 = n - (-n) = 2n \]

4. **Calculate the area of the rhombus**:
The area \( A \) of a rhombus can be calculated using the formula:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
Substituting the values of \( d_1 \) and \( d_2 \):
\[ A = \frac{1}{2} \times (2n) \times (2n) = \frac{1}{2} \times 4n^2 = 2n^2 \]

5. **Set the area equal to 2 square units**:
Given that the area is 2 square units:
\[ 2n^2 = 2 \]
Dividing both sides by 2:
\[ n^2 = 1 \]

### Conclusion:
The condition we found is:
\[ n^2 = 1 \]