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    One consideration must be made before you help me with the solution. I am a layman. If you find it hard to explain it to a layman, just imagine that I am your mum (assuming, of course, that your mum is entirely mathematically illiterate)

    I can't figure out how anything we have learnt in C3 can be applied to the first part of the question.

    Jan 07 OCR(MEI) Core 3

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    I would really appreciate the help or at least some sympathy
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    (Original post by Nwtr)
    One consideration must be made before you help me with the solution. I am a layman. If you find it hard to explain it to a layman, just imagine that I am your mum (assuming, of course, that your mum is entirely mathematically illiterate)

    I can't figure out how anything we have learnt in C3 can be applied to the first part of the question.

    I would really appreciate the help or at least some sympathy
    This doesn't require any C3 specific knowledge, only GCSE knowledge.

    All you need to do is find the distance AQ in terms of y (as requested), then note the relation AQ+AP = 6 so AP = 6-AQ hence you can then determine AP in terms of y. So, now you know the hypotenuse of the triangle OAP in terms of y.

    Now note that, alternatively, you can express the hypotenuse of the triangle in terms of x by considering Pythagoras' Theorem, and the fact that the legs of the triangle are of lengths x and 3 respectively.

    Equate the two expressions for the hypotenuse to yield the equality shown.
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    (Original post by Nwtr)
    I would really appreciate the help or at least some sympathy
    Wouldn't expect a layman to be doing C3, so ....

    To get you started:

    i) AQ= AO - OQ

    and just plug in the lengths.


    AP+AQ= 6, so AP=...

    The equation arises from considering Pythagoras on AOP

    Can you take it from there? If not, please be specific with the issue you're having.

    Note: MEI questions tend to be more challenging/interesting than the other boards.
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    (Original post by RDKGames)
    This doesn't require any C3 specific knowledge, only GCSE knowledge.

    All you need to do is find the distance AQ in terms of y (as requested), then note the relation AQ+AP = 6 so AP = 6-AQ hence you can then determine AP in terms of y. So, now you know the hypotenuse of the triangle OAP in terms of y.

    Now note that, alternatively, you can express the hypotenuse of the triangle in terms of x by considering Pythagoras' Theorem, and the fact that the legs of the triangle are of lengths x and 3 respectively.

    Equate the two expressions for the hypotenuse to yield the equality shown.
    Yeah, I knew it was something to do with vectors. I am just having a little trouble knowing how to express Q(0,y) A(0,3) in terms of y. All that is going through my mind is 3-y

    Thank you for your help.
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    (Original post by Nwtr)
    Yeah, I knew it was something to do with vectors. I am just having a little trouble knowing how to express Q(0,y) A(0,3) in terms of y. All that is going through my mind is 3-y
    You got it then.
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    (Original post by RDKGames)
    You got it then.
    Ah, thank you very much for taking the time to help. Have a great day!
 
 
 
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