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A curve is traced by a point $P(x,y)$ which moves such that its distance from the point $A(-1,1)$ is three times its distance from the point $B(2,-1)$. Determine the equation of the curve.

I have only one question. And that is the only thing I need answered at this time. My question to you is, when it says "which moves such that its distance from the point..." by distance, does it mean the slope from $P(x,y)$ to $A(-1,1)$ is three times the distance than from$P(x,y)$ to $B(2,-1)$? Please answer only this and nothing else. I will re-edit with further findings.

Edit: To find the next points would it be logical to use this equation:

$d=distance$ and $P=(x,y)$ $$d(P,(-1,1))=3d(P,(2,-1))$$ $$d\sqrt{(-1\pm x_1)^2+(1\pm y_1)^2}=3d\sqrt{(2\pm x_1)^2+(-1\pm y_1)^2}$$ And from here I would use $P=(x,y)$ and plug in any values of $x$ and $y$ to try and find my equation. Would this be correct?

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    You don't need the $d$ any more in the square roots. Remove the $d$, square both sides of your equation, and simplify accordingly.2012-07-24
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    I would prefer for simplicity $\sqrt{x+1)^2+(y-1)^2}=3\sqrt{(x-2)^2+(y+1)^2}$. Equivalently and much better, square both sides, bring stuff together, there will be a nice surprise.2012-07-24
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    @AndréNicolas: I think this is right, but I have no idea. I got; $-\dfrac{1}{2}x-4y-\dfrac{11}{4}=0$2012-07-24
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    Sorry, dinner event. Start from my comment above, square both sides we get (x+1)^2+(y-1)^2=9[(x-2)^2+ 9(y+1)^2]$. Expand and simplify. You wil get the equation of a **circle**. (It will start as $8x^2+8y^2+\cdots=0$. Divide by $8$ to make it look more familiar.2012-07-25
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    @AndréNicolas sorry I was on a little vacation. Came back to this problem and I got the equation: $0=8(x^2-y^2-\frac{1}{4}x+2y+5\frac{1}{4})$. I do not believe this is correct in any way.2012-08-04
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    You are right, there are some errors in your calculation. Expand. We get $x^2+2x+1+y^2-2y+1=9x^2-36x+36+9y^2+18y+9$, which partly simplifies to $8x^2+8y^2-38x+20y+43=0$. To make this look even more familiar, but messier, divide everything by $8$. By completing the square, you can identify the centre and radius of this circle.2012-08-04
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    Okay, so I got: $8\left((x^2-\frac{19}{4}x)(y^2+\frac{5}{2}y)+\frac{43}{8}\right)$ and then complete the square?2012-08-07

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