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

Answers (1)
D Divya Prakash Singh

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.

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