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 value of \( n \) given that the area of the rhombus formed by the lines \( x \pm y \pm n = 0 \) is equal to 2 square units.

### Step-by-Step Solution:

1. **Identify the Lines**:
The lines given are:
- \( x + y + n = 0 \) (Line 1)
- \( x + y - n = 0 \) (Line 2)
- \( x - y + n = 0 \) (Line 3)
- \( x - y - n = 0 \) (Line 4)

2. **Find the Intersection Points**:
To find the vertices of the rhombus, we need to find the intersection points of these lines:
- Intersection of Line 1 and Line 3:
\[ \begin{align*} x + y + n &= 0 \\ x - y + n &= 0 \end{align*} \]
Adding these equations:
\[ 2x + 2n = 0 \implies x = -n \]
Substituting \( x = -n \) in Line 1:
\[ -n + y + n = 0 \implies y = 0 \]
So, one vertex is \( (-n, 0) \).

- Intersection of Line 1 and Line 4:
\[ \begin{align*} x + y + n &= 0 \\ x - y - n &= 0 \end{align*} \]
Adding these equations:
\[ 2x = 0 \implies x = 0 \]
Substituting \( x = 0 \) in Line 1:
\[ 0 + y + n = 0 \implies y = -n \]
So, another vertex is \( (0, -n) \).

- Intersection of Line 2 and Line 3:
\[ \begin{align*} x + y - n &= 0 \\ x - y + n &= 0 \end{align*} \]
Adding these equations:
\[ 2x = n \implies x = \frac{n}{2} \]
Substituting \( x = \frac{n}{2} \) in Line 2:
\[ \frac{n}{2} + y - n = 0 \implies y = \frac{n}{2} \]
So, another vertex is \( \left( \frac{n}{2}, \frac{n}{2} \right) \).

- Intersection of Line 2 and Line 4:
\[ \begin{align*} x + y - n &= 0 \\ x - y - n &= 0 \end{align*} \]
Adding these equations:
\[ 2x - n = 0 \implies x = \frac{n}{2} \]
Substituting \( x = \frac{n}{2} \) in Line 2:
\[ \frac{n}{2} + y - n = 0 \implies y = -\frac{n}{2} \]
So, the last vertex is \( \left( \frac{n}{2}, -\frac{n}{2} \right) \).

3. **Calculate the Length of the Diagonals**:
The diagonals of the rhombus are:
- From \( (-n, 0) \) to \( (0, -n) \): Length = \( \sqrt{(-n - 0)^2 + (0 + n)^2} = \sqrt{n^2 + n^2} = n\sqrt{2} \)
- From \( (0, n) \) to \( (n, 0) \): Length = \( \sqrt{(0 - n)^2 + (n - 0)^2} = n\sqrt{2} \)

4. **Area of the Rhombus**:
The area \( A \) of a rhombus is given by:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
Where \( d_1 \) and \( d_2 \) are the lengths of the diagonals. Here both diagonals are \( n\sqrt{2} \):
\[ A = \frac{1}{2} \times n\sqrt{2} \times n\sqrt{2} = \frac{1}{2} \times 2n^2 = n^2 \]

5. **Set the Area Equal to 2**:
Given that the area is 2 square units:
\[ n^2 = 2 \]

### Conclusion:
Thus, the value of \( n^2 \) is \( 2 \).