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If the point $ P(x, y) $ is equidistant from the points $ A(6,1) $ and $ B(1,6) $, find the relation between $ x $ and $ y $.
Given:
The point $P (x, y)$ is equidistant from the points $A (6, 1)$ and $B (1, 6)$.
To do:
We have to find the relation between \( x \) and \( y \).
Solution:
$P (x, y)$ is equidistant from the points $A (6, 1)$ and $B (1,6)$. This implies,
\( \mathrm{PA}=\mathrm{PB} \)
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{PA}=\sqrt{(6-x)^{2}+(1-y)^{2}} \)
\( \mathrm{PB}=\sqrt{(1-x)^{2}+(6-y)^{2}} \)
This implies,
\( \sqrt{(6-x)^{2}+(1-y)^{2}}=\sqrt{(1-x)^{2}+(6-y)^{2}} \)
Squaring on both sides, we get,
\( (6-x)^{2}+(1-y)^{2}=(1-x)^{2}+(6-y)^{2} \)
\( 36+x^{2}-12 x+1+y^{2}-2 y=1+x^{2}-2 x+36 +y^{2}-12 y \)
\( -12 x-2 y=-2 x-12 y \)
\( -12 x+2 x=-12 y+2 y \)
\( -10 x=-10 y \)
\( x=y \)
Therefore, the relation between \( x \) and \( y \) is $x=y$.