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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."


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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
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
@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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