If the points (x, y) be equidistant from the points a(a+b, b-a) and b(a-b, a+b), prove that bx-ay=0

Solution:-

It is given that the points (x,y) be equidistant from the points A(a+b,b-a) and B(a-b,a+b).

PA=PB

Take square both side,

PA^2=PB^2

Now use distance

formula ,

{x-(a+b)}^2+{y-(b-a)}^2={x-(a-b)}^2+{y-(a+b)}^2

=>x^2+(a+b)^2-2x(a+b)+y^2+(b-a)^2-2y(b-a)y=x^2+(a-b)^2-2x(a-b)+y^2+(a+b)^2-2y(a+b)

=>2x(a-b)-2x(a+b)=2y(b-a)-2y(a+b)

=>2x{a-b-a-b}=2y{b-a-a-b}

=>2x(-2b)=2y(-2a)

=>bx=ay

Hence, it is proved.

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