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Math Help - Geometry: True or False

  1. #1
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    Geometry: True or False

    Need help with this, if it is true or false and why.

    "If a circle with its centre at the origin has a chord with slope m, the equation of the right bisector of the chord is "y=mx".
    Is this statement true or false, and can you give me detailed reasons for your answer.

    I have been re reading my notes and going thru my books and I can not figure this out at all.
    Can someone help me on this please, and thanks for your reply.
    Joanne
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  2. #2
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    Hello, Joanne!

    Did you make a sketch?


    If a circle with its centre at the origin has a chord with slope m,
    the equation of the right bisector of the chord is: y \,=\,mx

    Is this statement true or false? . . . . false
    Code:
                 A  |
                  o * *
              *     *     *
            *       | *     *
           *        |   *    *
                    |   * *
          *         | *     * *
      - - * - - - - * - - - - o - -
          *         |         * B
                    |
           *        |        *
            *       |       *
              *     |     *
                  * * *
                    |
    Chord AB has slope m.

    The perpendicular bisector of the chord does pass through the origin,
    . . but its slope is -\frac{1}{m}

    Its equation is: . y \:=\:-\frac{1}{m}\,y

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  3. #3
    Senior Member mollymcf2009's Avatar
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    A chord that is a perpendicular bisector of another chord would have the negative reciprocal slope of the other chord.

    If a chord has slope = m, its perpendicular bisector will have the slope = -\frac{1}{m}
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  4. #4
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    Why does the circle matter? I guess it helps to talk about the "chord".

    Slope of chord is m.
    Supposed slope of right bisector is m, from y = mx
    Supposed slope of right bisector, perpendicular to chord, is -1/m.

    Is m = -1/m?

    Exploration: Look up the "Normal Form" of a line. Interesting stuff.

    No, wait! The circle keeps the chord away from the Origin. Why does that matter, or does it? More exploration.
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  5. #5
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    Quote Originally Posted by bradycat View Post
    Need help with this, if it is true or false and why.

    "If a circle with its centre at the origin has a chord with slope m, the equation of the right bisector of the chord is "y=mx".
    Is this statement true or false, and can you give me detailed reasons for your answer.

    I have been re reading my notes and going thru my books and I can not figure this out at all.
    Can someone help me on this please, and thanks for your reply.
    Joanne
    Chords are straight lines. Straight lines have equation y = mx+C. You're trying to prove that, in our case C = 0.

    The circle is centred at the origin, hence any line which passes through its centre will pass through the y axis at y = 0, hence C = 0. All chords of a circle pass through the centre.
    Last edited by mr fantastic; January 30th 2009 at 07:11 PM. Reason: No edit - just flagging the reply as having been moved from a duplicate thread.
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  6. #6
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    Quote Originally Posted by Mush View Post
    Chords are straight lines.
    Line segments.

    All chords of a circle pass through the centre.
    A diameter does. Did you mean that all perpendicular bisectors of chords pass through the centre?
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