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If the point $P( x,\ y)$ is equidistant from the points $A( a\ +\ b,\ b\ β\ a)$ and $B( a\ β\ b,\ a\ +\ b)$. Prove that $bx=ay$.
Given: Point $P( x,\ y)$ is equidistant from the points $A( a\ +\ b,\ b\ –\ a)$ and $B( a\ –\ b,\ a\ +\ b)$
To do: To Prove that $bx=ay$.
Solution:
$P( x,\ y)$ is equidistant from the points $A( a\ +\ b,\ b\ –\ a)$ and $B( a\ –\ b,\ a\ +\ b)$.
we know if there two points $( x_{1} ,\ y_{1})$ and $( x_{2} ,\ y_{2})$,
Distance between the two points,$=\sqrt{( x_{2} -x_{1})^{2} +( y_{2} -y_{1})^{2}}$
By using the distance formula,
$PA=\sqrt{( a+b-x)^{2} +( b-a-y)^{2}}$
Similarly, $PB=\sqrt{( a-b-x)^{2} +( a+b-y)^{2}}$
Because, $PA=PB$
$\Rightarrow \sqrt{( a+b-x)^{2} +( b-a-y)^{2}} =\sqrt{( a-b-x)^{2} +( a+b-y)^{2}}$
$\Rightarrow ( a+b-x)^{2} +( b-a-y)^{2} =( a-b-x)^{2} +( a+b-y)^{2}$
$\Rightarrow ( a+b)^{2} +x^{2} -2( a+b) x+( b-a)^{2} +y^{2} -2( b-a) y=( a-b)^{2} +x^{2} -2( a-b) x+( a+b)^{2} +y^{2} -2( a+b) y$
$\Rightarrow -2( a+b) x-2( b-a) y=-2( a-b) x-2( a+b) y$
$\Rightarrow -2ax-2bx-2by+2ay=-2ax+2bx-2ay-2by$
$\Rightarrow -2bx+2ay=2bx-2ay$
$\Rightarrow 4bx=4ay$
$\Rightarrow bx=ay$
$\therefore bx=ay ....( proved)$
Updated on: 10-Oct-2022
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