If the point ( x,y ) is equidistant form the points ( a+b,b-a ) and (a-b ,a+b ) , prove that bx = ay
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Solution
As per the question, we have
`sqrt((x-a-b)^2 +(y-b+a)^2 ) = sqrt((x-a+b)^2 +(y-a-b)^2)`
`⇒(x-a-b)^2 +(y-b+a)^2 = (x-a+b)^2 +(y-a-b)^2` (Squaring both sides)
`⇒x^2 + (a+b)^2 -2x (a+b) +y^2 +(a-b)^2 -2y(a-b)=x^2 +(a-b)^2 -2x(a-b)+y^2 +(a+b)^2 -2y (a+b)`
`⇒-x(a+b) - y (a-b) = -x(a-b) -y(a+b)`
`⇒-xa -xb -ay +by = -xa + bx -ya-by`
⇒ by=bx
Hence, . bx = ay
Concept: Coordinate Geometry
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