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  • Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

10.  Find a relation between x and y such that the point (x, y) is equidistant from the point  (3, 6) and (– 3, 4).

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Let the point P(x,y ) is equidistant from A(3,6) and B(-3,4).

Then, the distances AP =BP

AP = \sqrt{(x-3)^2+(y-6)^2}  and  BP = \sqrt{(x-(-3))^2+(y-4)^2}

\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}

Squaring both sides: we obtain

\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2

\Rightarrow (2x)(-6)+(2y-10)(-2)= 0        \left [\because a^2-b^2 = (a+b)(a-b) \right ]

\Rightarrow -12x-4y+20 = 0

\Rightarrow 3x+y-5 = 0

Thus, the relation is  3x+y-5 = 0 between x and y.

Posted by

Divya Prakash Singh

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