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(D) For a rectangular hyperbola,the asymptotes are mutually perpendicular. Given one asymptote is $3x - 4y - 6 = 0$,the other must be of the form $4x

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$4x + 3y + 17 = 0$

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(D) For a rectangular hyperbola,the asymptotes are mutually perpendicular. Given one asymptote is $3x - 4y - 6 = 0$,the other must be of the form $4x + 3y + \lambda = 0$.The centre of the hyperbola is the intersection point of the two asymptotes.Solving $3x - 4y - 6 = 0$ and $4x + 3y + \lambda = 0$:Multiply the first by $3$ and the second by $4$: $9x - 12y - 18 = 0$ and $16x + 12y + 4\lambda = 0$.Adding these gives $25x + 4\lambda - 18 = 0$,so $x = \frac{18 - 4\lambda}{25}$.Substituting $x$ into $3x - 4y - 6 = 0$: $4y = 3(\frac{18 - 4\lambda}{25}) - 6 = \frac{54 - 12\lambda - 150}{25} = \frac{-96 - 12\lambda}{25}$,so $y = \frac{-24 - 3\lambda}{25}$.Since the centre lies on $x - y - 1 = 0$,we have $\frac{18 - 4\lambda}{25} - (\frac{-24 - 3\lambda}{25}) - 1 = 0$.$18 - 4\lambda + 24 + 3\lambda - 25 = 0$ $\Rightarrow 17 - \lambda = 0$ $\Rightarrow \lambda = 17$.Thus,the equation of the other asymptote is $4x + 3y + 17 = 0$.

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

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