Question

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

Answer 1

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 proved

Answer 2

Distance between the points P(x, y) and  A(a+b, b-a) & B(a-b, a+b) are equal  

Therefore AP=BP

i.e,

√{[x - (a + b)]2 + [y - (b -a)]}2 = √{x - (a - b)]2 + [y - (a + b)]2}

x2 + (a + b)2 - 2x(a + b) + y2 + (b - a)2 - 2y(b - a) = x2 + (a - b)2 - 2x(a - b) + y2 + (a + b)2 - 2y(a + b)

-2ax - 2bx - 2by + 2ay = - 2ax + 2bx - 2ay - 2by

ay - bx = bx - ay

2ay = 2bx

bx = ay

Hence proved.