Find the relation between x and y, if the point (x, y) is equidistant from (7, -6),(-3, 4)

Tuesday, December 31, 2013

Find the relation between x and y, if the point (x, y) is equidistant from (7, -6),(-3, 4)

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Problem: How to Find the relation between x and y, if the point (x, y)is equidistant from (7, -6),(-3, 4).

$(7,-6),(-3,4)\,$

Solutions:

Find the distance between (x,y) and the first point is

$\sqrt{(x-7)^2+(y+6)^2}\,$

$\sqrt{x^2+49-14x+y^2+36+12y}\,$

$\sqrt{x^2+y^2-14x+12y+85}\,$

Let this expression be 1.

Find the distance between (x,y) and the second point is

$\sqrt{(x+3)^2+(y-4)^2}\,$

$\sqrt{x^2+9+6x+y^2+16-8y}\,$

$\sqrt{x^2+y^2+6x-8y+25}\,$

Let this expression be 2.

As per the given condition,equating both the expressions,

$\sqrt{x^2+y^2-14x-12y+85}=\sqrt{x^2+y^2+6x-8y+25}\,$

Squaring on both sides,we get

$x^2+y^2-14x+12y+22=x^2+y^2+6x-8y+25\,$

cancelling the terms which are common on both sides,

$-14x-6x+12y+8y=-60\,$

$-20x+20y=-60\,$

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Tags: distance, Geometry Problems

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Tuesday, December 31, 2013