The equation of the line passing through the centre of a rectangular hyperbola is `x-y-1=0`. If one of its asymptotoes is `3x-4y-6=0`, the equation of the other asymptote is

Answer

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

### Step 1: Identify the given equations
We have the following equations:
1. The equation of the line passing through the center of the hyperbola:
\[ x - y - 1 = 0 \quad \text{(Equation L1)} \]
2. The equation of one asymptote:
\[ 3x - 4y - 6 = 0 \quad \text{(Asymptote A1)} \]

### Step 2: Find the point of intersection of L1 and A1
To find the intersection point, we can solve the equations simultaneously. First, we rewrite L1:
\[ y = x - 1 \]
Now, substitute \(y\) in the equation of A1:
\[ 3x - 4(x - 1) - 6 = 0 \]
Expanding this gives:
\[ 3x - 4x + 4 - 6 = 0 \]
Simplifying:
\[ -x - 2 = 0 \implies x = -2 \]
Now, substitute \(x = -2\) back into L1 to find \(y\):
\[ y = -2 - 1 = -3 \]
Thus, the point of intersection is:
\[ (-2, -3) \]

### Step 3: Find the slope of the first asymptote A1
The equation of A1 can be rewritten in slope-intercept form:
\[ 3x - 4y - 6 = 0 \implies 4y = 3x - 6 \implies y = \frac{3}{4}x - \frac{3}{2} \]
The slope \(m_1\) of A1 is:
\[ m_1 = \frac{3}{4} \]

### Step 4: Find the slope of the second asymptote A2
Since the asymptotes of a rectangular hyperbola are perpendicular, the product of their slopes is -1:
\[ m_1 \cdot m_2 = -1 \implies \frac{3}{4} \cdot m_2 = -1 \]
Solving for \(m_2\):
\[ m_2 = -\frac{4}{3} \]

### Step 5: Write the equation of the second asymptote A2
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m_2 = -\frac{4}{3}\) and the point \((-2, -3)\):
\[ y + 3 = -\frac{4}{3}(x + 2) \]
Multiplying through by 3 to eliminate the fraction:
\[ 3(y + 3) = -4(x + 2) \]
Expanding:
\[ 3y + 9 = -4x - 8 \]
Rearranging gives:
\[ 4x + 3y + 17 = 0 \]

### Final Equation
The equation of the other asymptote is:
\[ 4x + 3y + 17 = 0 \]