Question

I want the answer with (with formula) clarity ​

Answer 1

GIVEN :-

$\sf \: A \: point \: (x,y) \: is \: in \: equidistant \: from \: two \\ \sf\: points \: (a+b , b-a) \: and \: (a-b , a+b )$

TO PROOF :-

$\rm \: bx \: = \: ay$

PROOF :-

$\sf \: Distance \: formula = \sqrt{( x_{2} -x_{1} ) {}^{2} - ( y_{2} -y_{1} ) {}^{2} }$

$\sf \:1 \star \: condition$

$\sf \: x_{1} = x \: \: \: \: \: \: \: \: \: \: y_{1} = y \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: x_{2} = (a + b . \: b - a) \: \: \: \: \: \: \: \: \: \: y_{2} = (a - b \: .a + b)$

$\sf \: Distance \: formula = \sqrt{( x_{2} -x_{1} ) {}^{2} - ( y_{2} -y_{1} ) {}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf = \sqrt{(a + b - x) {}^{2} - (b - a - y) {}^{2} } \: \: \: .......(1)$

$\sf \: similarly \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: 2 \star \: condition \:$

$\sf \: Distance \: formula = \sqrt{( x_{2} -x_{1} ) {}^{2} - ( y_{2} -y_{1} ) {}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf = \sqrt{(a - b - x) {}^{2} - (a + b - y) {}^{2} } \: \: \: .......(2)$

$\sf \: According \: \: to \: \: the \: \: question$

$\sf \: distance \: \: of \: \: (x.y) \: from \: \: given \: point \: are \: equal$

$\sf \sqrt{(a + b - x) {}^{2} - (b - a - y) {}^{2} } = \sqrt{(a - b - x) {}^{2} - (a + b - y) {}^{2} } \\ \\ \sf \: squaring \: \: both \: sides \\ \\ \sf \implies\: ( {a + b - x)}^{2} - ( {b - a - y)}^{2} = (a - b - x) {}^{2} - (a + b - y) {}^{2} \\ \\ \sf \: appling \: ( {x}^{2} - {y}^{2} ) = (x + y)(x - y) \\ \\ \implies\sf\big[a + b - x + b - a - y]\big[a + b - x - b + a + y] = \big[a - b - x + a + b - y]\big[a - b - x - a - b + y] \\ \\ \sf \implies \: (2b - x - y)(2a - x + y) = (2a - x - y)( - 2b - x + y)$