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    How do I do part d? Any hints on how to start?
    Thanks
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    (Original post by Super199)
    How do I do part d? Any hints on how to start?
    Thanks
    Just sub p into the equation, if it is = to 0 then it satisfies.
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    (Original post by zed963)
    Just sub p into the equation, if it is = to 0 then it satisfies.
    What equation?
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    (Original post by Super199)
    What equation?
    p^2-4p-16=0
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    (Original post by zed963)
    p^2-4p-16=0
    I don't know the value of p .
    I am given AC. Where A is (2,5) and c is (p, some y coordinate) which is equal to 5? It's worth 4 marks so I dunno
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    (Original post by Super199)
    I don't know the value of p .
    I am given AC. Where A is (2,5) and c is (p, some y coordinate) which is equal to 5? It's worth 4 marks so I dunno
    It clearly says " has x coordinate which is equal to p"

    Post your working out to the question.
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    What is the coordinates of point C in terms of p?

    What is the length of AC in terms of p?

    Answer those and you should be able to solve the rest.
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    (Original post by zed963)
    Just sub p into the equation, if it is = to 0 then it satisfies.
    You cannot do this, you don't know what p is yet.
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    (Original post by james22)
    You cannot do this, you don't know what p is yet.
    I assumed the OP already had P.
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    (Original post by james22)
    What is the coordinates of point C in terms of p?

    What is the length of AC in terms of p?

    Answer those and you should be able to solve the rest.
    If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
    Then for AC I would say
    AC^2=(p-2)^2+(y coordinate -5)^2?
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    (Original post by Super199)
    If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
    Then for AC I would say
    AC^2=(p-2)^2+(y coordinate -5)^2?
    Name:  ImageUploadedByStudent Room1397673804.420931.jpg
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    I've tried doing it but its not as easy as it looks.


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    (Original post by Super199)
    If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
    Then for AC I would say
    AC^2=(p-2)^2+(y coordinate -5)^2?
    You know what the y coordinate will be in terms of p because C lies on the line and you worked out the equation of the line earlier in the question.
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    (Original post by davros)
    You know what the y coordinate will be in terms of p because C lies on the line and you worked out the equation of the line earlier in the question.
    Oh wow yh hang on I will have another go. Thanks
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    I have the solution to this? Want hints?
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    C= (P,-1/2P+6)
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    You then plug this into the AC^2 formula
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    (Original post by zed963)
    I assumed the OP already had P.
    RTFQ
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    (Original post by Galileo Galilei)
    C= (P,-1/2P+6)
    Can you explain where you've got this from?
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    (Original post by james22)
    RTFQ
    Sorry guys.
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    (Original post by zed963)
    Can you explain where you've got this from?
    Mathematical logic
 
 
 
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