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If the points $(2, 1)$ and $(1, -2)$ are equidistant from the point $(x, y)$, show that $x + 3y = 0$.
Given:
The points $(2, 1)$ and $(1, -2)$ are equidistant from the point $(x, y)$.
To do:
We have to show that $x + 3y = 0$.
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 $(2,1)$ and \( (x, y) \) \( =\sqrt{(x-2)^{2}+(y-1)^{2}} \)
The distance between $(1,-2)$ and \( (x, y) \) \( =\sqrt{(x-1)^{2}+(y+2)^{2}} \)
The points $(2,1)$ and $(1,-2)$ are equidistant from \( (x, y) \)
\( \therefore \sqrt{(x-2)^{2}+(y-1)^{2}}=\sqrt{(x-1)^{2}+(y+2)^{2}} \)
Squaring on both sides, we get,
\( \Rightarrow(x-2)^{2}+(y-1)^{2}=(x-1)^{2}+(y+2)^{2} \)
\( \Rightarrow x^{2}-4 x+4+y^{2}-2 y+1=x^{2}-2 x+1+y^{2}+4y+4 \)
\( \Rightarrow x^{2}-4 x+y^{2}-2 y+5-x^{2}+2 x-y^{2}-4 y-5=0 \)
\( \Rightarrow-2 x-6 y=0 \)
\( \Rightarrow -2(x+3 y)=0 \)
\( \Rightarrow x+3 y=0 \)
Hence proved.
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