If the area of the rhombus enclosed by lines `lx+-my+-n=0` be 2 square units, then:

Options

A. `m^(2)=n`
B. `n^(2)=lm`
C. `m=ln`
D. `n-ln`

Answer

To solve the problem, we need to find the relationship between the coefficients of the lines given by the equation \( lx \pm my \pm n = 0 \) and the area of the rhombus formed by these lines.

### Step-by-Step Solution:

1. **Identify the Lines**:
The lines given are \( lx - my - n = 0 \), \( lx - my + n = 0 \), \( lx + my - n = 0 \), and \( lx + my + n = 0 \). These lines will intersect to form a rhombus.

2. **Find the Vertices of the Rhombus**:
To find the vertices of the rhombus, we can set \( y = 0 \) in the equations:
- For \( lx - my - n = 0 \): \( x = \frac{n}{l} \)
- For \( lx - my + n = 0 \): \( x = -\frac{n}{l} \)
- The vertices on the x-axis are \( \left(-\frac{n}{l}, 0\right) \) and \( \left(\frac{n}{l}, 0\right) \).

Now, set \( x = 0 \):
- For \( lx + my - n = 0 \): \( y = \frac{n}{m} \)
- For \( lx + my + n = 0 \): \( y = -\frac{n}{m} \)
- The vertices on the y-axis are \( \left(0, -\frac{n}{m}\right) \) and \( \left(0, \frac{n}{m}\right) \).

3. **Calculate the Lengths of the Diagonals**:
The lengths of the diagonals \( D_1 \) and \( D_2 \) of the rhombus are:
- \( D_1 = \frac{2n}{l} \) (distance between the points on the x-axis)
- \( D_2 = \frac{2n}{m} \) (distance between the points on the y-axis)

4. **Area of the Rhombus**:
The area \( A \) of the rhombus can be calculated using the formula:
\[ A = \frac{1}{2} \times D_1 \times D_2 \]
Substituting the values of \( D_1 \) and \( D_2 \):
\[ A = \frac{1}{2} \times \left(\frac{2n}{l}\right) \times \left(\frac{2n}{m}\right) = \frac{2n^2}{lm} \]

5. **Set the Area Equal to 2**:
We are given that the area of the rhombus is 2 square units:
\[ \frac{2n^2}{lm} = 2 \]
Simplifying this equation:
\[ n^2 = lm \]

### Conclusion:
Thus, the relationship we find is:
\[ n^2 = lm \]