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    Hello everyone

    I don't know if I'm being mindless as it is so late but for the life of me I do not know the method to find B in this question.

    It does not give me an equation.

    Can anyone help me?

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    (Original post by grahamcrackers)
    Hello everyone

    I don't know if I'm being mindless as it is so late but for the life of me I do not know the method to find B in this question.

    It does not give me an equation.

    Can anyone help me?

    drop a perpendicular from the centre to the x axis
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    (Original post by TeeEm)
    drop a perpendicular from the centre to the x axis
    Thanks for the reply!

    I tried doing that, but I got stuck.
    Do I form a triangle?
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    (Original post by grahamcrackers)
    Thanks for the reply!

    I tried doing that, but I got stuck.
    Do I form a triangle?
    Yep. Think about the properties of triangle ABC
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    (Original post by grahamcrackers)
    Thanks for the reply!

    I tried doing that, but I got stuck.
    Do I form a triangle?
    what is the x coordinate of the point where your perpendicular meets the x axis?
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    (Original post by TeeEm)
    what is the x coordinate of the point where your perpendicular meets the x axis?
    I got B to be (7, 0)

    I knew that A to C was moved by the vector across 3 and up to, so when I reached the centre I went down 2 and across 3.

    Is this right? If so, I feel like this is a stupid way of doing it. Is there a nicer way?
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    (Original post by grahamcrackers)
    I got B to be (7, 0)

    I knew that A to C was moved by the vector across 3 and up to, so when I reached the centre I went down 2 and across 3.

    Is this right? If so, I feel like this is a stupid way of doing it. Is there a nicer way?
    you are correct

    nicer I am not sure ...

    there is a circle theorem which states that the perpendicular bisector of a chord always passes though the centre.

    the chord here is the x axis
    bisector cuts x axis at (4,0)
    as this is the midpoint (perpendicular bisector) the other end must be at (7,0)
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    (Original post by TeeEm)
    you are correct

    nicer I am not sure ...

    there is a circle theorem which states that the perpendicular bisector of a chord always passes though the centre.

    the chord here is the a axis
    bisector cuts x axis at (4,0)
    as this is the midpoint (perpendicular bisector) the other end must be at (7,0)
    Ah!! I understand! This method is much easier.

    Thank you for your time! I need to figure out how to give you rep! (i'm new here!)
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    (Original post by grahamcrackers)
    Ah!! I understand! This method is much easier.

    Thank you for your time! I need to figure out how to give you rep! (i'm new here!)
    my pleasure
 
 
 
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