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    Greetings, I've been baffled with a question for quite a while now and I seek help.

    A circle with its centre at (2,1) has the line x+2y=9 as a tangent.

    (a) Find the equation of a line through the centre, perpendicular to this tangent.



    Thank you!
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    (Original post by Nightly)
    Greetings, I've been baffled with a question for quite a while now and I seek help.

    A circle with its centre at (2,1) has the line x+2y=9 as a tangent.

    (a) Find the equation of a line through the centre, perpendicular to this tangent.

    Thank you!
    What is the gradient of the line x+2y=9? What is the negative reciprocal of it?

    Then just apply y-y_1=m(x-x_1)
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    (Original post by Nightly)
    Greetings, I've been baffled with a question for quite a while now and I seek help.

    A circle with its centre at (2,1) has the line x+2y=9 as a tangent.

    (a) Find the equation of a line through the centre, perpendicular to this tangent.
    There may some confusion, as you're given more information than necessary.

    The question asks you to find a line through the point (2,1) perpendicular to the line x+2y=9. The circle is irrelevant, at least for (a).
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    (Original post by RDKGames)
    What is the gradient of the line x+2y=9? What is the negative reciprocal of it?

    Then just apply y-y_1=m(x-x_1)
    The gradient in this case is 1 and so the negative reciprocal is -1. When I substitute into that equation, I get y = -x +3
    However, my book disagrees and says y = 2x +3 but I don't understand how.
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    (Original post by Nightly)
    The gradient in this case is 1 and so the negative reciprocal is -1. When I substitute into that equation, I get y = -x +3
    However, my book disagrees and says y = 2x +3 but I don't understand how.
    The gradient is not 1.

    You need to put the equation in the form y=mx+c first.


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    (Original post by RDKGames)
    The gradient is not 1.

    You need to put the equation in the form y=mx+c first.


    Posted from TSR Mobile
    I'm honestly ashamed I missed out on this. Thank you very much, I got the answer I needed.
 
 
 
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