Yiuyu
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Q: Find a formula for the following situation describing the set of possible location of the point Pj in cartesian coordinates
i) The distance from P1 to the point (1,3) is twice that from P1 to (-1,2)
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Yiuyu
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Is this what I am doing so far right?
(x-1)^2 + (y-3)^2 = 2((x+1)^2 + (y-2)^2)
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RDKGames
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(Original post by Yiuyu)
Is this what I am doing so far right?
(x-1)^2 + (y-3)^2 = 2((x+1)^2 + (y-2)^2)
Not quite.

The distance d_1 from P_1 to (1,3) is given by (x-1)^2+(y-3)^2=d_1^2

Then distance d_2 from P_1 to (-1,2) is given by (x+1)^2+(y-2)^2=d_2^2

We know that d_1=2d_2

Carry on from there.
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Yiuyu
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(Original post by RDKGames)
Not quite.

The distance d_1 from P_1 to (1,3) is given by (x-1)^2+(y-3)^2=d_1^2

Then distance d_2 from P_1 to (-1,2) is given by (x+1)^2+(y-2)^2=d_2^2

We know that d_1=2d_2

Carry on from there.
With every example will I have to square root?
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Yiuyu
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(Original post by RDKGames)
Not quite.

The distance d_1 from P_1 to (1,3) is given by (x-1)^2+(y-3)^2=d_1^2

Then distance d_2 from P_1 to (-1,2) is given by (x+1)^2+(y-2)^2=d_2^2

We know that d_1=2d_2

Carry on from there.
Thats basically the same thing as I did though except square rooting?
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RDKGames
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(Original post by Yiuyu)
With every example will I have to square root?
Not sure what you mean. I would avoid square roots.
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Yiuyu
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(Original post by RDKGames)
Not sure what you mean. I would avoid square roots.
So I expanded d1^2= x^2-2x+y^2-6y+10
and expanded d2^2=x^2+2x+y^2-4y+5
What do I do next?
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RDKGames
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(Original post by Yiuyu)
So I expanded d1^2= x^2-2x+y^2-6y+10
and expanded d2^2=x^2+2x+y^2-4y+5
What do I do next?
Merge the two somehow, using the fact how the two distances are related to each other.

You are aiming to end up with some sort of equation in terms of x and y (which is what is meant by Cartesian coordinates in your OP) and no d involved.
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Yiuyu
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(Original post by RDKGames)
Merge the two somehow, using the fact how the two distances are related to each other.

You are aiming to end up with some sort of equation in terms of x and y (which is what is meant by Cartesian coordinates in your OP) and no d involved.
since d1=2d2 right can we do d1^2=4d2^2 and then replace the d1^2 with the 4d2^2?
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RDKGames
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(Original post by Yiuyu)
since d1=2d2 right can we do d1^2=4d2^2 and then replace the d1^2 with the 4d2^2?
Yep.
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Yiuyu
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(Original post by RDKGames)
Yep.
What about this Q
The distance from P4 to the point (-5,3) is twice the distance from P4 to the line with equation 4x+3y+5=0
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Yiuyu
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#12
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Notnek
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Yiuyu
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(Original post by Yiuyu)
What about this Q
The distance from P4 to the point (-5,3) is twice the distance from P4 to the line with equation 4x+3y+5=0
RDKGames
Well Ik d1=2d2
which means d1^2=4d2^2
but how do i get d2?
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RDKGames
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(Original post by Yiuyu)
RDKGames
Well Ik d1=2d2
which means d1^2=4d2^2
but how do i get d2?
A point on the line has coordinates (x,-\frac{1}{3}(4x+5))
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Yiuyu
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(Original post by RDKGames)
A point on the line has coordinates (x,-\frac{1}{3}(4x+5))
Is that d2?
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RDKGames
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(Original post by Yiuyu)
Is that d2?
No, that's a point on the line.

d_2 would be the distance between P_1 and this point.

Though of course, since we are saying P_1 has coordinates (x,y), it would be helpful to instead say that the point on the line is in fact (p,-\frac{1}{3}(4p+5)) for some p \in \mathbb{R}
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Yiuyu
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(Original post by RDKGames)
No, that's a point on the line.

d_2 would be the distance between P_1 and this point.

Though of course, since we are saying P_1 has coordinates (x,y), it would be helpful to instead say that the point on the line is in fact (p,-\frac{1}{3}(4p+5)) for some p \in \mathbb{R}
Ohh okayy yup! May I have a hint how to work out d2? Ive worked out d1 but dont know how to work out d2
Btw thank you so much for helping me
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RDKGames
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(Original post by Yiuyu)
Ohh okayy yup! May I have a hint how to work out d2? Ive worked out d1 but dont know how to work out d2
Btw thank you so much for helping me
Same method to work out d_2, just that you'd have your equation with a p term.
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Yiuyu
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(Original post by RDKGames)
Same method to work out d_2, just that you'd have your equation with a p term.
Well the distance d_1 from P_1 to (-5,3) is given by (x+5)^2+(y-3)^2=d_1^2

I dont get how I'd use the same method? Like how am I supposed to get the values?
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RDKGames
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(Original post by Yiuyu)
Well the distance d_1 from P_1 to (-5,3) is given by (x+5)^2+(y-3)^2=d_1^2

I dont get how I'd use the same method? Like how am I supposed to get the values?
Well you're looking for the distance d_2 from P_1 to (p,-\frac{1}{3}(4p+5)) which is given by...?
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