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Just wondering how you would solve this:

"Find the locus of a point $P(x,y)$ which moves such that its distance from the $x$-axis is always one more unit than its distance from the $y$-axis."

Thanks

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do you see what is the "distance" (positive quantity) between a point $P(x,y)$ and $x-axis$... How do you define a point $(4,5)$ in co ordinate plane??? –  Praphulla Koushik Sep 7 '13 at 11:05
    
@yanbo I think that (homework) tag should only be added by the OP or if the OP explicitly says somewhere that it is homework. See meta. Here's the link to the revision history. –  Martin Sleziak Sep 7 '13 at 11:27
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1 Answer 1

up vote 2 down vote accepted

The perpendicular distance of point $(x,y)$ from the $x$ axis is $|y|$ and the distance from the $y$ axis is $|x|$.

So, $|y|=|x|+1$.

For $y\geq0$, the graph is $y=|x|+1$ and for $y<0$, the graph is given by $y=-|x|-1$.

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Hi, thanks for the help, but I'm unsure whether to use perpendicular distance to find the locus, or the normal distance formula? –  missiledragon Sep 7 '13 at 12:04
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@missiledragon The phrase, "the distance from the $x$ axis" most likely refers to the perpendicular distance from it. Any other distance measurement would require a point on the $x$ and $y$ axes from which the measurement is to be made using the distance formula between two points. –  Alraxite Sep 7 '13 at 12:13
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