$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{hyperbola} \mathrm{passing} \mathrm{through} \mathrm{the} \mathrm{origin} \mathrm{has} 3\mathrm{x}-4\mathrm{y}-1=0 \mathrm{and} 4\mathrm{x}-3\mathrm{y}-6=0 \mathrm{as} \mathrm{its} \mathrm{asymptotes}, \mathrm{then} \mathrm{the} \\ \mathrm{equation} \mathrm{of} \mathrm{its} \mathrm{transverse} \mathrm{and} \mathrm{conjugate} \mathrm{axes} \mathrm{are} \end{gathered}$$

The transverse axis is the bisector of the angle between asymptotes containing the origin and the conjugate axis is the other bisector. So, their equations are given by 

$$\begin{gathered} \frac{3x-4y-1}{\sqrt{25}}=\frac{4x-3y-6}{\sqrt{25}} and \frac{\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{-}\mathbf{1}}{\sqrt{\mathbf{25}}}\mathbf{=}\frac{\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{6}\mathbf{)}}{\sqrt{\mathbf{25}}}\mathbf{ } \\ \mathbf{\Rightarrow }\mathbf{3}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathbf{3}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{-}\mathbf{6}\mathbf{=}\mathbf{0} \\ \mathbf{\Rightarrow }\mathit{x}\mathbf{+}\mathit{y}\mathbf{-}\mathbf{5}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{x}\mathbf{-}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0} \end{gathered}$$