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    From the FP2 textbook, it's asking for a sketch of the locus and it's Cartesian equation for z.

     \frac{\left | z+3i \right |}{\left | z-6i \right |} = 1

    The textbook reckons I'm wrong, and the Cartesian equation is:
     y=-0.5x+9/4

    Can anyone point out where I've gone wrong? I couldn't find any examples of a division question, so maybe it's wrong for me to multiply by  |z-6i| ?
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    (Original post by alexhazmat)
    From the FP2 textbook, it's asking for a sketch of the locus and it's Cartesian equation for z.

     \frac{\left | z+3i \right |}{\left | z-6i \right |} = 1

    The textbook reckons I'm wrong, and the Cartesian equation is:
     y=-0.5x+9/4

    Can anyone point out where I've gone wrong? I couldn't find any examples of a division question, so maybe it's wrong for me to multiply by  |z-6i| ?
    are you sure the numerator is z+3i and not z+3, or the equivalent for the denominator?


    based off the top of my head and on the actual answer I'd guess that's where the problem is
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    (Original post by ValerieKR)
    are you sure the numerator is z+3i and not z+3, or the equivalent for the denominator?


    based off the top of my head and on the actual answer I'd guess that's where the problem is
    I was thinking that, but its definitely z+3i :Name:  9d935a4484944aa650d9f68f37df915f.png
Views: 75
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    (Original post by alexhazmat)
    I was thinking that, but its definitely z+3i :Name:  9d935a4484944aa650d9f68f37df915f.png
Views: 75
Size:  4.2 KB
    Then it's a typing mistake in the textbook/worksheet/whatever

    That provided answer is an answer to a question where that i in the numerator isn't real (yay imaginary number jokes)
    Your solution is correct for the given question
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    (Original post by ValerieKR)
    Then it's a typing mistake in the textbook/worksheet/whatever

    That provided answer is an answer to a question where that i in the numerator isn't real (yay imaginary number jokes)
    Your solution is correct for the question provided
    Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors
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    (Original post by alexhazmat)
    Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors
    btw there is a much quicker way to do those ones

    once you have mod(z-a)=mod(z-b)
    The equation is the equation of the perpendicular bisector of AB (because the distance from A and B is equal)
    (and if A=B it's just everywhere)
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    (Original post by ValerieKR)
    btw there is a much quicker way to do those ones

    once you have mod(z-a)=mod(z-b)
    The equation is the equation of the perpendicular bisector of AB (because the distance from A and B is equal)
    (and if A=B it's just everywhere)
    I knew it was a bisector of AB, I just like doing it the long way to show a bit of reasoning.

    However I don't understand why, for example:  |z-6| = 2|z+6-9i| ends up being a circle, instead of a bisector?

    Algebraically I see that it ends up being a circle, but I don't see why (the textbook fails at explaining why doubling the 2nd part completely changes it from a bisector to a circle)
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    (Original post by alexhazmat)
    I knew it was a bisector of AB, I just like doing it the long way to show a bit of reasoning.

    However I don't understand why, for example:  |z-6| = 2|z+6-9i| ends up being a circle, instead of a bisector?

    Algebraically I see that it ends up being a circle, but I don't see why (the textbook fails at explaining why doubling the 2nd part completely changes it from a bisector to a circle)
    Because of the '2' - it's always twice as far away from one point as another, so you end up with a solution on one side of one complex number, and on the other

    think
    -----a------b------
    if we mark on every point equidistant from a and b we get
    -----a---x---b------

    and every one twice as far from a as b
    -----a----x--b----x--

    it's an extension of that extra solution appearing to 2D

    the factor not being 1 destroys the symmetry in a sense and allows solutions to appear on either side of either a or b on that 1D line
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    (Original post by ValerieKR)
    Because of the '2' - it's always twice as far away from one point as another, so you end up with a solution on one side of one complex number, and on the other

    think
    -----a------b------
    if we mark on every point equidistant from a and b we get
    -----a---x---b------

    and every one twice as far from a as b
    -----a----x--b----x--

    it's an extension of that extra solution appearing to 2D

    the factor not being 1 destroys the symmetry in a sense and allows solutions to appear on either side of either a or b on that 1D line
    Ahhhh that makes sense now. I never thought of it that way, initially I thought the coordinates of B would be doubled. Thanks for the insight!
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    (Original post by alexhazmat)
    Okay thanks, I knew I wasn't losing my mind. Sadly it seems these textbooks are riddled with errors
    That is why they are free.
    I had similar issues
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    (Original post by IYGB)
    That is why they are free.
    I had similar issues
    They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)
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    (Original post by alexhazmat)
    They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)
    Bloody hell - I know for certain you can get resources for this type of question of the AQA website for free
    and I suspect a lot of other places too
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    (Original post by alexhazmat)
    They're free? Each book is £16 per book on amazon. I'm only complaining because these books are my only source of practice questions (apart from past papers)
    It used to be free on the AQA website
 
 
 
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