Question

find a relation between x and y such that point (x,y) is equidistant from the points (3,6) and (-3,4)

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{euidistant}\mathrm{from}\mathrm{A}\left(3,6\right)\mathrm{and}\mathrm{B}\left(-3,4\right). \\ \mathrm{now},\mathrm{PA}=\mathrm{PB} \\ \Rightarrow \sqrt{{\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-6\right)}^2}=\sqrt{{\left(\mathrm{x}+3\right)}^2+{\left(\mathrm{y}-4\right)}^2} \\ \Rightarrow {\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-6\right)}^2={\left(\mathrm{x}+3\right)}^2+{\left(\mathrm{y}-4\right)}^2 \\ \Rightarrow {\mathrm{x}}^2+9-6\mathrm{x}+{\mathrm{y}}^2+36-12\mathrm{y}={\mathrm{x}}^2+9+6\mathrm{x}+{\mathrm{y}}^2+16-8\mathrm{y} \\ \Rightarrow -6\mathrm{x}-12\mathrm{y}+45=6\mathrm{x}-8\mathrm{y}+25 \\ \Rightarrow 12\mathrm{x}+4\mathrm{y}-20=0 \\ \Rightarrow 3\mathrm{x}+\mathrm{y}-5=0 \\ \boxed{\mathbf{USING}\mathbf{}\mathbf{DISTANCE}\mathbf{}\mathbf{FORMULA}\mathbf{}\mathbf{:} \\ \mathbf{Distance}\mathbf{}\mathbf{between}\mathbf{}\left({\mathbf{x}}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\right)\mathbf{}\mathbf{and}\mathbf{}\left({\mathbf{x}}_{\mathbf{2}}\mathbf{,}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\right)\mathbf{}\mathbf{=}\mathbf{}\sqrt{{\left({\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\right)}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\left({\mathbf{y}}_{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{y}}_{\mathbf{1}}\right)}^{\mathbf{2}}}} \end{gathered}$$