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A point $P(x, y)$ is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance
from the line $x-y=1$. Then the equation of the locus of $P$ is
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from the line $x-y=1$. Then the equation of the locus of $P$ is
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Verified Answer
The correct answer is:
$x^2+y^2+2 x y+2 x-2 y-1=0$
It is given that the sum of squares of distance of point $P(x, y)$ is equal to the square of its distance from the line $x-y=1$, so
$$
\begin{aligned}
& x^2+y^2=\frac{(x-y-1)^2}{2} \\
\Rightarrow \quad 2 x^2+2 y^2 & =x^2+y^2+1-2 x y-2 x+2 y \\
\Rightarrow \quad & x^2+y^2+2 x y+2 x-2 y-1=0
\end{aligned}
$$
$$
\begin{aligned}
& x^2+y^2=\frac{(x-y-1)^2}{2} \\
\Rightarrow \quad 2 x^2+2 y^2 & =x^2+y^2+1-2 x y-2 x+2 y \\
\Rightarrow \quad & x^2+y^2+2 x y+2 x-2 y-1=0
\end{aligned}
$$
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