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Circles and gradients

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    http://www.thomas-reddington.com/upl...p_jan_2009.pdf

    For question 5a)

    In the mark scheme they are finding the gradeints for PQ and QR, then multiplying them, I have no idea why?
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    Find the gradient of PQ, find the perpendicular for the gradient of QR, find the equation of QR, substitute the y value for a.

    You should be familiar with m1m2 = -1, another way of finding the perpendicular gradient.
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    theyre perpandicular (because the triangle inside a semicircle is right angled) and hence the product of their gradients is -1.
    So from this it should be easy to work out your value for a
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    (Original post by Joshmeid)
    Find the gradient of PQ, find the perpendicular for the gradient of QR, find the equation of QR, substitute the y value for a.

    You should be familiar with m1m2 = -1, another way of finding the perpendicular gradient.
    This is the first time i've seen of m1m2 = -1.
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    (Original post by zoxe)
    This is the first time i've seen of m1m2 = -1.
    m1 is the gradient of PQ.

    Lets say PQ = 5 for example.

    we have m1m2 = -1

    We now substitute m1 = 5

    5m2 = -1

    m2 = -1/5
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    ive solved it with Pythagoras, however i'm still baffled on how theyve done it
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    (Original post by Joshmeid)
    m1 is the gradient of PQ.

    Lets say PQ = 5 for example.

    we have m1m2 = -1

    We now substitute m1 = 5

    5m2 = -1

    m2 = -1/5
    Why does PQ = gradient?

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