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A point moves in such a way that the sum of its distance from $x y$-plane and $y z$-plane remains equal to its distance from $z x$-plane. The locus of the point is
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Verified Answer
The correct answer is:
$x-y+z=0$
We know that,
Distance from \(x y\)-plane to the point \(P(x, y, z)=|z|\)
Distance from \(z x\)-plane to the point \(P(x, y, z)=|y|\)
Distance from \(y z\)-plane to the point \(P(x, y, z)=|x|\)
It is given that,
the sum of its distance from \(x y\)-plane and \(y z\)-plane remains equal to its distance from zx-plane
\(\begin{aligned}
& \Rightarrow|x|+|z|=|y| \\
& \Rightarrow x+z-y=0
\end{aligned}\)
Distance from \(x y\)-plane to the point \(P(x, y, z)=|z|\)
Distance from \(z x\)-plane to the point \(P(x, y, z)=|y|\)
Distance from \(y z\)-plane to the point \(P(x, y, z)=|x|\)
It is given that,
the sum of its distance from \(x y\)-plane and \(y z\)-plane remains equal to its distance from zx-plane
\(\begin{aligned}
& \Rightarrow|x|+|z|=|y| \\
& \Rightarrow x+z-y=0
\end{aligned}\)
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