Question

If the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$, prove that $bx=ay$.

Answer 1

Step 1: Calculate the distance $PA$

Given: $P\left(x,y\right)$ is equidistant from $A(a+b,b-a)$ and $B(a-b,a+b)$.

i.e., $PA=PB$

The distance $\left(d\right)$ between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ :

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ [ Distance formula ]

The distance between the points $P\left(x,y\right)$ and $A(a+b,b-a)$ is,

$$\begin{gathered} PA=\sqrt{{\left(a+b-x\right)}^2+{\left(b-a-y\right)}^2} \\ \Rightarrow PA=\sqrt{\left(a^2+b^2+x^2+2ab-2bx-2ax\right)+\left(b^2+a^2+y^2-2ab+2ay-2by\right)}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{+}\mathbf{2}\mathbf{b}\mathbf{c}\mathbf{+}\mathbf{2}\mathbf{c}\mathbf{a}\mathbf{]} \\ \Rightarrow PA=\sqrt{a^2+b^2+x^2+2ab-2bx-2ax+b^2+a^2+y^2-2ab+2ay-2by} \\ \Rightarrow PA=\sqrt{2a^2+2b^2+x^2+y^2-2bx-2ax+2ay-2by} \end{gathered}$$

Step 2: Calculate the distance $PB$

The distance between the points $P\left(x,y\right)$ and $B(a-b,a+b)$ is,

$$\begin{gathered} PB=\sqrt{{\left(a-b-x\right)}^2+{\left(a+b-y\right)}^2} \\ \Rightarrow PB=\sqrt{\left(a^2+b^2+x^2-2ab+2bx-2ax\right)+\left(a^2+b^2+y^2+2ab-2by-2ay\right)}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{+}\mathbf{2}\mathbf{b}\mathbf{c}\mathbf{+}\mathbf{2}\mathbf{c}\mathbf{a}\mathbf{]} \\ \Rightarrow PB=\sqrt{a^2+b^2+x^2-2ab+2bx-2ax+a^2+b^2+y^2+2ab-2by-2ay} \\ \Rightarrow PB=\sqrt{2a^2+2b^2+x^2+y^2+2bx-2ax-2by-2ay} \end{gathered}$$

Step 3: Prove the required equation

According to the question, the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$.

i.e., $PA=PB$

$\Rightarrow$ $\sqrt{2a^2+2b^2+x^2+y^2-2bx-2ax+2ay-2by}=\sqrt{2a^2+2b^2+x^2+y^2+2bx-2ax-2by-2ay}$

$\Rightarrow$ $2a^2+2b^2+x^2+y^2-2bx-2ax+2ay-2by=2a^2+2b^2+x^2+y^2+2bx-2ax-2by-2ay$ [Squaring on both sides]

$\Rightarrow$ $-2bx-2ax+2ay-2by=2bx-2ax-2by-2ay$

$\Rightarrow$ $-2bx+2ay=2bx-2ay$

$\Rightarrow$ $2ay+2ay=2bx+2bx$

$\Rightarrow$ $4ay=4bx$

$\Rightarrow$ $ay=bx$

Hence, it is proved that $bx=ay$.