Question

if the point P(x,y) is equidistant from the points Q(a+b, b-a) and R(a-b, a+b), then prove that bx = ay.

Answer 1

(a-b)²=(b-a) ²
I hope this is helpful

Answer 2

Hey Dear ! 

Since P(x,y) is equidistant from the point A(a-b, a+b) and B(a+b, b-a) 

PA = PB 

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

SBS 

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

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

Cancel x² + y² on both side 

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

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

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

⇒ 2x(a+b-a+b) = 2y(a+b-b+a)

⇒ 2x(2b) = 2y(2a)

⇒ bx = ay

Hence proved