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Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$.
Given:
The points $(3, 6)$ and $(-3, 4)$ are equidistant from the point $(x, y)$.
To do:
We have to find a relation between $x$ and $y$.
Solution:
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,
The distance between $(3,6)$ and \( (x, y) \) \( =\sqrt{(x-3)^{2}+(y-6)^{2}} \)
The distance between $(-3,4)$ and \( (x, y) \) \( =\sqrt{(x+3)^{2}+(y-4)^{2}} \)
The points $(3,6)$ and $(-3,4)$ are equidistant from \( (x, y) \).
\( \therefore \sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}} \)
Squaring on both sides, we get,
\( \Rightarrow(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2} \)
\( \Rightarrow x^{2}-6 x+9+y^{2}-12 y+36=x^{2}+6 x+9+y^{2}-8y+16 \)
\( \Rightarrow x^{2}-6 x+y^{2}-12 y+36-x^{2}-6 x-y^{2}+8 y-16=0 \)
\( \Rightarrow-12 x-4 y+20=0 \)
\( \Rightarrow -4(3x+ y-5)=0 \)
\( \Rightarrow 3x+y-5=0 \)
The required relation is $3x+y-5=0$.
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