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    Ques b.ii) and working (sorry i drew another diagram in pencil).
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    You know that the gradient of m_{\overrightarrow{AB}} = \frac{\Delta y}{\Delta x} = - \frac{3}{2}
    You know that the line connecting B and C is orthogonal to this line. Therefore m_{\overrightarrow{BC}}= \frac{2}{3}
    You can use this to find k
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    It's important to realise that point C will lie on the line that you gave in part bi)
    And the y coordinate of C is 7.
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    am I too late?
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    (Original post by TeeEm)
    am I too late?
    i think so just like me again ;(
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    (Original post by B_9710)
    It's important to realise that point C will lie on the line that you gave in part bi)
    And the y coordinate of C is 7.
    (Original post by MathsAndChess)
    You know that the gradient of m_{\overrightarrow{AB}} = \frac{\Delta y}{\Delta x} = - \frac{3}{2}
    You know that the line connecting B and C is orthogonal to this line. Therefore m_{\overrightarrow{BC}}= \frac{2}{3}
    You can use this to find k
    I subbed in 7 to get k = 5 but Im confused - how do you know c lies on this line?
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    (Original post by kiiten)
    I subbed in 7 to get k = 5 but Im confused - how do you know c lies on this line?
    Right angle = 90 degrees.

    Perpendicular = 90 degrees.

    Ringing any bells?
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    (Original post by kiiten)
    Ques b.ii) and working (sorry i drew another diagram in pencil).
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    draw a sketch and you'll simply spot that for the condition that ABC is 90 is must on lie on the line perpendicular to AB
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    (Original post by Zacken)
    Right angle = 90 degrees.

    Perpendicular = 90 degrees.

    Ringing any bells?
    Yes. Thanks

    What about finding the distance from a circle to a tangent at a point (outside the circle). Would you just find the distance from centre to the point then minus the radius. - I got confused on the 'tangent' part.

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    (Original post by kiiten)
    Yes. Thanks

    What about finding the distance from a circle to a tangent at a point (outside the circle). Would you just find the distance from centre to the point then minus the radius. - I got confused on the 'tangent' part.

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    Sketch it, use the fact that the angle between the radius and the tangent at a point is 90 degrees, then use trigonometry or Pythagoras.
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    (Original post by Zacken)
    Sketch it, use the fact that the angle between the radius and the tangent at a point is 90 degrees, then use trigonometry or Pythagoras.
    Is it like this (finding the hypotenuse)? The length of PC is bigger than the radius and i dont think P lies on the circle.

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    (Original post by kiiten)
    Is it like this (finding the hypotenuse)? The length of PC is bigger than the radius and i dont think P lies on the circle.

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    Not quite, more like this:



    Except only one half is rquired and none of those lines inside the circle. Just the one line from the centre to the circumference (the radius line) and then the tangent to the point. The angle CAP is a right angle.
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    (Original post by Zacken)
    Not quite, more like this:



    Except only one half is rquired and none of those lines inside the circle. Just the one line from the centre to the circumference (the radius line) and then the tangent to the point. The angle CAP is a right angle.
    Ohh so the tangent could be either AP or BP?

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    (Original post by kiiten)
    Ohh so the tangent could be either AP or BP?

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    Yes, your sketch can be either one of those. I just couldn't find a proper picture that showed only one, block half of it our, sorry.
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    (Original post by Zacken)
    Yes, your sketch can be either one of those. I just couldn't find a proper picture that showed only one, block half of it our, sorry.
    Dw its fine, at least you made the effort. I understood what you were trying to say. Thank you!!

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