Question

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IF THE GIVEN POINT(X,Y) IS EQUIDISTANT FROM OTHER 2 POINTS Q(a+b,b-a)AND R (a-b,a+b) THEN PROVE THAT bx=ay.....

Answer 1

Distace between the points (x, y) and  (a+b, b-a) & (a-b, a+b) is equal  ⇒ √{[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.

Answer 2

Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR

⇒ $\sqrt{{x−(a+b)}² + {y−(b−a)}²} = \sqrt{{x−(a–b)}² + {y−(a+b)}²}$

⇒{x−(a+b)}² + {y−(b−a)}² = {x−(a−b)}² + {y−(a+b)}²

⇒x²−2x(a+b) + (a+b)² + y²−2y(b−a)+(b−a)² = x²+(a−b)² −2x(a−b)+y² −2y(a+b)+(a+b)²

⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)

⇒ax+bx+by−ay = ax−bx+ay+by

⇒2bx = 2ay

⇒bx = ay

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