Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (−3, 4).
The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula
$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Let the three given points be P(x, y), A(3,6) and B(−3,4).
Now let us find the distance between ‘P’ and ‘A’.
$P A=\sqrt{(x-3)^{2}+(y-6)^{2}}$
Now, let us find the distance between ‘P’ and ‘B’.
$P B=\sqrt{(x+3)^{2}+(y-4)^{2}}$
It is given that both these distances are equal. So, let us equate both the above equations,
$P A=P B$
$\sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}}$
Squaring on both sides of the equation we get,
$(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$
$x^{2}+9-6 x+y^{2}+36-12 y=x^{2}+9+6 x+y^{2}+16-8 y$
$12 x+4 y=20$
$3 x+y=5$
Hence the relationship between ' $x$ ' and ' $y$ ' based on the given condition is $3 x+y=5$.
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