Question

Find a relation between $x\mathrm{and}y$ such that the point $P\left(x,y\right)$ is equidistant from the points $A\left(-5,3\right)\mathrm{and}B\left(7,2\right)$.

Answer 1

Finding a relation between $x\mathrm{and}y$:

It is known that the distance $d$ between two points $\left(x_1,y_1\right)\mathrm{and}\left(x_2,y_2\right)$ can be found by using the formula given below:

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ …(1)

According to question, $PA=PB$.

$$\begin{gathered} \Rightarrow \sqrt{{\left(-5-x\right)}^2+{\left(3-y\right)}^2}=\sqrt{{\left(7-x\right)}^2+{\left(2-y\right)}^2} \\ \Rightarrow {\left(-5-x\right)}^2+{\left(3-y\right)}^2={\left(7-x\right)}^2+{\left(2-y\right)}^2\left[\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow 25+x^2+10x+9+y^2-6y=49+x^2-14x+4+y^2-4y \\ \Rightarrow 24x-2y=19 \end{gathered}$$

Final Answer:

Hence, the relation between $x\mathrm{and}y$ is $\mathbf{24}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{19}$.