Locus help (1 Viewer)

samsonhohoho · Mar 2, 2013

Find the equation of the path of a point P(x,y) as it moves in the XY-Plane such that its distance to the line y=x is always 1/2 its distance to the origin.

I know how the graph looks like, but i dont know how to find the equation. Please show working THANKS!!!

asianese · Mar 3, 2013

We must use the perpendicular distance formula because we are dealing with P, a point, and a line, x-y=0.

So, suppose the line is called $l$ and the origin is $O$ . Then:

$Pl=\frac{1}{2}\times PO\\ \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} = \frac{1}{2} \times \sqrt{x^2+y^2} $ since the distance of P to the origin is such.$ \\ $ Let's square both sides. $ \\ 4\times \frac{\left(1\times x-1\times y+0\right)^2}{1^2+1^2} = x^2+y^2.$

You should be able to go from there. I think you will need to impose a restriction on your final answer since we squared both sides.

Locus help (1 Viewer)

samsonhohoho

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asianese

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