find the equation of the line which pass through the point of intersection of the lines 4x-3y-1=0 & 2x-5y+3=0 and is equally inclined to the axes

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$$\begin{gathered} \mathrm{Let} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{be} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \\ 4\mathrm{x}-3\mathrm{y}-1=0 ;\mathrm{equation}\left(\mathrm{i}\right) \\ 2\mathrm{x}-5\mathrm{y}+3=0 ;\mathrm{equation}\left(\mathrm{ii}\right) \\ \mathrm{equation}\left(\mathrm{i}\right)-2\times \mathrm{equation}\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}, \\ 4\mathrm{x}-3\mathrm{y}-1-\left(4\mathrm{x}-10\mathrm{y}+6\right)=0 \\ \Rightarrow 4\mathrm{x}-3\mathrm{y}-1-4\mathrm{x}+10\mathrm{y}-6=0 \\ \Rightarrow 7\mathrm{y}=7 \\ \Rightarrow \mathrm{y}=1 \\ \mathrm{Pu} \mathrm{this} \mathrm{value} \mathrm{of} \mathrm{y} \mathrm{in} \mathrm{equation}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ 4\mathrm{x}-3-1=0 \\ \Rightarrow \mathrm{x}=1 \\ \Rightarrow \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,1\right) \\ \mathrm{Now} \mathrm{since} \mathrm{required} \mathrm{line} \mathrm{is} \mathrm{equally} \mathrm{inclined} \mathrm{to} \mathrm{axes}. \\ \Rightarrow \mathrm{Angle} \mathrm{made} \mathrm{by} \mathrm{line} \mathrm{with} \mathrm{positive} \mathrm{direction} \mathrm{of} \mathrm{x}-\mathrm{axis} \mathrm{may} \mathrm{be} 45° \mathrm{or} 135°. \\ \mathrm{Case} \left(\mathrm{i}\right) \\ \mathrm{m}=\tan 45° \\ \Rightarrow \mathrm{m}=1 \\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{is} \mathrm{given} \mathrm{by}, \\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \Rightarrow \mathrm{y}-1=1\left(\mathrm{x}-1\right) \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathrm{Case} \left(\mathrm{ii}\right) \\ \mathrm{m}=\tan 135° \\ \Rightarrow \mathrm{m}=-1 \\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{is} \mathrm{given} \mathrm{by}, \\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \Rightarrow \mathrm{y}-1=-1\left(\mathrm{x}-1\right) \\ \Rightarrow \mathrm{y}-1=-\mathrm{x}+1 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{2} \\ \mathbf{Hence}\mathbf{ }\mathbf{required}\mathbf{ }\mathbf{equations}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{line}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{.} \end{gathered}$$

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