if 2x+y-4=0 is a bisector of angles between the lines a(x-1) + b(y-2) =0,c (x-1) + d(y-2)= 0 the other angular bisector is?

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{a}\left(\mathrm{x}-1\right)+\mathrm{b}\left(\mathrm{y}-2\right)=0....\left(\mathrm{i}\right) \\ \mathrm{c}\left(\mathrm{x}-1\right)+\mathrm{d}\left(\mathrm{y}-2\right)=0....\left(\mathrm{ii}\right) \\ \mathrm{Clearly} \left(1,2\right) \mathrm{lie} \mathrm{on} \mathrm{both} \mathrm{the} \mathrm{lines} \mathrm{and} \mathrm{hence} \left(1,2\right) \mathrm{is} \mathrm{their} \mathrm{point} \mathrm{of} \mathrm{intersection}. \\ \mathrm{Both} \mathrm{the} \mathrm{bisectors} \mathrm{will} \mathrm{also} \mathrm{pass} \mathrm{through} \left(1,2\right) \\ \mathrm{One} \mathrm{of} \mathrm{the} \mathrm{bisector} \mathrm{is} 2\mathrm{x}+\mathrm{y}-4=0 \\ \mathrm{Other} \mathrm{bisector} \mathrm{will} \mathrm{be} \mathrm{perpendicular} \mathrm{to} \mathrm{this} \mathrm{bisector}. \mathrm{Hence} \mathrm{its} \mathrm{equation} \mathrm{will} \mathrm{be} \\ \mathrm{x}-2\mathrm{y}=\mathrm{\lambda } \\ \mathrm{It} \mathrm{passes} \mathrm{through} \left(1,2\right) \\ \Rightarrow 1-4=\mathrm{\lambda } \\ \Rightarrow \mathrm{\lambda }=-3 \\ \mathrm{Hence} \mathrm{equation} \mathrm{is} \\ \mathrm{x}-2\mathrm{y}=-3 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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