# Geometry

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#1
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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#2
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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2 years ago
#3
(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 from to is given by

Then distance from to is given by

We know that

Carry on from there.
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#4
(Original post by RDKGames)
Not quite.

The distance from to is given by

Then distance from to is given by

We know that

Carry on from there.
With every example will I have to square root?
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#5
(Original post by RDKGames)
Not quite.

The distance from to is given by

Then distance from to is given by

We know that

Carry on from there.
Thats basically the same thing as I did though except square rooting?
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2 years ago
#6
(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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#7
(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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2 years ago
#8
(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 and (which is what is meant by Cartesian coordinates in your OP) and no involved.
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#9
(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 and (which is what is meant by Cartesian coordinates in your OP) and no 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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2 years ago
#10
(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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#11
(Original post by RDKGames)
Yep.
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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#12
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#13
(Original post by Yiuyu)
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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2 years ago
#14
(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
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#15
(Original post by RDKGames)
A point on the line has coordinates
Is that d2?
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2 years ago
#16
(Original post by Yiuyu)
Is that d2?
No, that's a point on the line.

would be the distance between and this point.

Though of course, since we are saying has coordinates , it would be helpful to instead say that the point on the line is in fact for some
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#17
(Original post by RDKGames)
No, that's a point on the line.

would be the distance between and this point.

Though of course, since we are saying has coordinates , it would be helpful to instead say that the point on the line is in fact for some
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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2 years ago
#18
(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 , just that you'd have your equation with a term.
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#19
(Original post by RDKGames)
Same method to work out , just that you'd have your equation with a term.
Well the distance from to is given by

I dont get how I'd use the same method? Like how am I supposed to get the values?
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2 years ago
#20
(Original post by Yiuyu)
Well the distance from to is given by

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 from to which is given by...?
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