The line `2x + y = 0` passes through the centre of a rectangular hyperbola, one of whose asymptotes is `x - y = 1.` The equation of the other asymptotes is

Options

A. `3x + 3y + 1=0`
B. `3x + 3y -1 =0`
C. `x - 2y =0`
D. none of these

Answer

To find the equation of the other asymptote of the rectangular hyperbola, we can follow these steps:

### Step 1: Identify the given information
We have:
1. The equation of one asymptote: \( x - y = 1 \)
2. The equation of the line passing through the center of the hyperbola: \( 2x + y = 0 \)

### Step 2: Determine the slope of the given asymptote
The equation of the asymptote \( x - y = 1 \) can be rewritten in slope-intercept form:
\[ y = x - 1 \]
From this, we see that the slope \( m_1 \) of the first asymptote is \( 1 \).

### Step 3: Find the slope of the other asymptote
For a rectangular hyperbola, the asymptotes are perpendicular to each other. Therefore, if the slope of one asymptote is \( m_1 \), the slope of the other asymptote \( m_2 \) can be found using the relationship:
\[ m_1 \cdot m_2 = -1 \]
Substituting \( m_1 = 1 \):
\[ 1 \cdot m_2 = -1 \implies m_2 = -1 \]

### Step 4: Find the intersection point of the two lines
We need to find the intersection of the lines \( 2x + y = 0 \) and \( x - y = 1 \). We can solve these equations simultaneously.

From the first equation, we can express \( y \) in terms of \( x \):
\[ y = -2x \]

Now substitute \( y \) in the second equation:
\[ x - (-2x) = 1 \implies x + 2x = 1 \implies 3x = 1 \implies x = \frac{1}{3} \]

Now, substitute \( x = \frac{1}{3} \) back into the equation for \( y \):
\[ y = -2\left(\frac{1}{3}\right) = -\frac{2}{3} \]

Thus, the intersection point is \( \left(\frac{1}{3}, -\frac{2}{3}\right) \).

### Step 5: Write the equation of the other asymptote
Now that we have the slope of the other asymptote \( m_2 = -1 \) and the point \( \left(\frac{1}{3}, -\frac{2}{3}\right) \), we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = \left(\frac{1}{3}, -\frac{2}{3}\right) \) and \( m = -1 \):
\[ y - \left(-\frac{2}{3}\right) = -1\left(x - \frac{1}{3}\right) \]
This simplifies to:
\[ y + \frac{2}{3} = -x + \frac{1}{3} \]
Rearranging gives:
\[ y + x = \frac{1}{3} - \frac{2}{3} \implies y + x = -\frac{1}{3} \]

### Final Answer
The equation of the other asymptote is:
\[ x + y = -\frac{1}{3} \]