Question

Find the equation of the locus of the point which moves so that it's distance from the line 3x-4y+2=0 is equal to it's distance from the origin.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{point}\mathrm{is}\left(\mathrm{h},\mathrm{k}\right), \\ \mathrm{Now}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \mathrm{Distance}\mathrm{from}\mathrm{origin}=\mathrm{distance}\mathrm{between}\left(\mathrm{h},\mathrm{k}\right)\mathrm{and}3\mathrm{x}-4\mathrm{y}+2=0 \\ \sqrt{{\left(\mathrm{h}-0\right)}^2+{\left(\mathrm{k}-0\right)}^2}=\left|\frac{3\mathrm{h}-4\mathrm{k}+2}{\sqrt{3^2+{\left(-4\right)}^2}}\right|\left[\mathrm{Using}\mathrm{distance}\mathrm{formula}\mathrm{p}=\left|\frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right|\right] \\ \mathrm{Squaring}\mathrm{we}\mathrm{get}, \\ {\mathrm{h}}^2+{\mathrm{k}}^2=\frac{3\mathrm{h}-4\mathrm{k}+2}{5} \\ \mathrm{Now}\mathrm{replace}\mathrm{h}\mathrm{to}\mathrm{x}\mathrm{and}\mathrm{k}\mathrm{to}\mathrm{y}\mathrm{we}\mathrm{get}, \\ 5{\mathrm{x}}^2+5{\mathrm{y}}^2-3\mathrm{x}+4\mathrm{y}-2=0 \end{gathered}$$