Question

Find a relation between $x$ and $y$ such that the point $\left(x,y\right)$ is equidistant from the point $\left(3,6\right)$ and $\left(-3,4\right)$.

Answer 1

Step $1$: Define the relation between $x$ and $y$.

Let point, $P=\left(x,y\right)$ , $Q=\left(3,6\right)$ and $R=\left(-3,4\right).$

By using the distance formula, $d=\sqrt{{(x_2-x_1)}^2+{\left(y_2-y_1\right)}^2}$

Step $2$: Define the distance $PQ$ and $PR$.

$PQ=\sqrt{{\left(x-3\right)}^2+{\left(y-6\right)}^2}$ (Applying distance formula)

$PR=\sqrt{{\left(x+3\right)}^2+{\left(y-4\right)}^2}$ (Applying distance formula)

Now, according to the question $PQ=PR$ , therefore,

Step $3$: Equate $PQ$ and $PR$.

$$\begin{gathered} \Rightarrow \sqrt{{\left(x-3\right)}^2+{\left(y-6\right)}^2}=\sqrt{{\left(x+3\right)}^2+{\left(y-4\right)}^2} \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-6\right)}^2={\left(x+3\right)}^2+{\left(y-4\right)}^2 \\ \Rightarrow x^2+9-6x+y^2+36-12y=x^2+9+6x+y^2+16-8y \\ \Rightarrow -6x-12y-6x+8y=25-45 \\ \Rightarrow -12x-4y=-20 \\ \Rightarrow 3x+y=5 \end{gathered}$$

Hence, the relation between $x$ and $y$ is $3x+y=5$.