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Math Help - Right Bisector of a Chord?

  1. #1
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    Right Bisector of a Chord?

    Right Bisector of a Chord?

    The equation of a circle with centre O (0,0) is x2 +y2 = 169
    the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB


    1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint
    2) Find the slopes of AB-- igot 3/2 for slope of AB
    **3) Find Equation of the Line DE --- idont noe how to do this
    a) Use the negative reciprocal slope of AB for Slope of DE
    b) Use point slope form to find equation of DE


    igot help for numebr 3

    and igot y= (-2/3)x +b

    idont get how they got (-2/3)
    can yu explain tht process
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by ahmads121 View Post
    Right Bisector of a Chord?

    The equation of a circle with centre O (0,0) is x2 +y2 = 169
    the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB


    1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint How can a line segment have two midpoints?! Remember the formula:
    Midpoint(m)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}  {2}\right)
    2) Find the slopes of AB-- igot 3/2 for slope of AB Very good.
    **3) Find Equation of the Line DE --- idont noe how to do this Use the midpoint that you obtain in 1. to find the equation using the point slope form of a line.
    a) Use the negative reciprocal slope of AB for Slope of DE. Because the slope of a perpindicular line will be the negative reciprocal of the line it is perpindicular to.
    b) Use point slope form to find equation of DE. Done.


    igot help for numebr 3

    and igot y= (-2/3)x +b

    idont get how they got (-2/3) Draw 2 perpindicular lines at the origin and analyze their slopes. You will be enlightened.
    can yu explain tht process
    ..
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  3. #3
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    Quote Originally Posted by ahmads121 View Post
    Right Bisector of a Chord?

    The equation of a circle with centre O (0,0) is x2 +y2 = 169
    the points A(0,13) and B(-12,-5) are endpoints of the chord AB. DE right bisects chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB


    1) Find midpoint of AB -- igot (-6,0) and (0,4) for midpoint
    You mean that the midpoint is (-6, 4).

    2) Find the slopes of AB-- igot 3/2 for slope of AB
    **3) Find Equation of the Line DE --- idont noe how to do this
    The equation of a line through point (a, b) with slope m is y= m(x- a)+ b.
    Here, this is y= (3/2)(x- (-6))+ 4 or y= (3/2)x+ 9+ 4= (3/2)x+ 13.
    You don't actually need the equation of this line to answer the question.

    a) Use the negative reciprocal slope of AB for Slope of DE
    b) Use point slope form to find equation of DE


    igot help for numebr 3

    and igot y= (-2/3)x +b

    idont get how they got (-2/3)
    can yu explain tht process
    If two lihes, given by y= m_1 x+ b_1 and y= m_2x+ b_2 are perpendicular then m_1\times m_2= -1. Here m_1= \frac{3}{2} so you have \frac{3}{2} m_2= -1. Multiplying both sides by 2, 3m_2= -2. Dividing both sides by 3, m_2= -\frac{2}{3}.

    Now, you want that line to go through the midpoint of AB which was (-6, 4) so you must have 4= (-2/3)(-6)+ b= 4+ b. Adding b to both sides, b= 0.
    The equation of the perpendicular to the chord through its midpoint is y= (-2/3).
    You could also have used the formula I gave before: the equation of the line through (-6, 4) with slope -2/3 is y= -2/3(x-(-6))+ 4= -2/3(x+ 6)+ 4= -(2/3)x- 4+ 4= (-2/3) x again. Now do you see that the origin, (0, 0), satisfies that equation?
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