Maths / Coordinate Geometry / Distance Formula

QUESTION

 If the point (x, y) is equidistant from (a +b, b-a) and (a-b, a+b), show that bx =ay.
EXPLANATION
Explain TypeExplanation Content
Text

Let C =(x,y) A=(a + b, b-a) and B= (a-b, a+b)

$\large AC = \sqrt{(x-(a+b))^2+(y-(b-a))^2}$

$\large BC = \sqrt{(x- (a-b))^2+(y-(a+b)^2}$

As C is equidistant from A and B

AC= BC

$\large \sqrt{(x-(a+b))^2+(y-(b-a))^2} =\sqrt{(x- (a-b))^2+(y-(a+b)^2}$

Squaring Both sides

$\large (x-(a+b))^2+(y-(b-a))^2 =(x- (a-b))^2+(y-(a+b)^2$

$\large (x-a-b)^2+(y-b+a)^2 =(x- a+b)^2+(y-a-b)^2$

$\large (x-a-b)^2 -(x- a+b)^2 =(y-a-b)^2-(y-b+a)^2$

$\large (x-a-b)+(x- a+b) (x-a-b)-(x- a+b)$

$\large =((y-a-b)+(y-b+a))((y-a-b)-(y-b+a))$

$\large (2x-2a) (-2b) =(2y-2b)(-2a)$

$\large (-4xb+ 4ab) =(4ay+4ab)$

$\large xb =ay$

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