Question

Find the equation of the line equidistant from the point (2,-2) and the line 3x-4y+1=0

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{line}\mathrm{which}\mathrm{is}\mathrm{equidistant}\mathrm{from}\mathrm{point}(2,-2)\mathrm{and}3\mathrm{x}-4\mathrm{y}+1=0\mathrm{mmust}\mathrm{be}\mathrm{parallel}\mathrm{to} \\ 3\mathrm{x}-4\mathrm{y}+1=0. \\ \mathrm{i},\mathrm{e},\mathrm{equation}\mathrm{of}\mathrm{reqd}\mathrm{line}\mathrm{is}3\mathrm{x}-4\mathrm{y}+\mathrm{c}=0 \\ \mathrm{Now}, \\ \mathrm{perpendicular}\mathrm{distance}\mathrm{from}(2,-2)=\frac{\left|\mathrm{Ax}+\mathrm{By}+\mathrm{c}\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}=\frac{\left|3\times 2-4(-2)+\mathrm{c}\right|}{\sqrt{3^2+4^2}}=\frac{\left|14+\mathrm{c}\right|}{5} \\ \mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{lines}=\frac{\left|{\mathrm{c}}_1-{\mathrm{c}}_2\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}=\frac{\left|1-\mathrm{c}\right|}{5} \\ \mathrm{Now},\frac{\left|14+\mathrm{c}\right|}{5}=\frac{\left|1-\mathrm{c}\right|}{5} \\ \Rightarrow 14+\mathrm{c}=1-\mathrm{c} \\ \Rightarrow \mathrm{c}=\frac{-13}{2} \\ \mathrm{Thus},\mathrm{required}\mathrm{line}:3\mathrm{x}-4\mathrm{y}-\frac{13}{2}=0 \end{gathered}$$