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Math Help - quadratic equation

  1. #1
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    quadratic equation

    hi i'm struggling with this:

    when a parabola crosses the x-axis at the points A and B, the axis of symmetry is the perpendicular bisector of the line joining the points A and B. since A is the point (-5, 0) and B is the point (2/3, 0) the equation of the line of symmetry is x = -13/6

    i know that the mid point is -13/6 but why is that the equation? i thought you had to go through the whole y - y1 = m(x - x1) business. when the gradient of a horizontal line is 0 how do you come to the conclusion that the equation is x = -13/6.

    thanks for any help
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  2. #2
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    Quote Originally Posted by mark View Post
    when a parabola crosses the x-axis at the points A and B, the axis of symmetry is the perpendicular bisector of the line joining the points A and B. since A is the point (-5, 0) and B is the point (2/3, 0) the equation of the line of symmetry is x = -13/6

    i know that the mid point is -13/6 but why is that the equation?
    Because the axis of symmetry is vertical. Vertical lines are x=k.
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  3. #3
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    i'm sorry but i don't really understand what you mean, i literally took up maths about 2 weeks ago. what would the equation of the horizontal line be and how would you figure that out? to find the perpendicular bisector i know you need to reverse the gradient around ie m1 * m2 = -1 but i can't see how you'd do anything when the gradient is 0. why do you not use the y - y1 = m(x - x1) can you explain in detail please how and why its all done?

    thanks

    thanks
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  4. #4
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    Quote Originally Posted by mark View Post
    what would the equation of the horizontal line be and how would you figure that out?
    Vertical lines have this form: x=k.
    Horizontal lines have this form: y=k.

    Examples.
    A horizontal line through (2,-3) is y=-3.
    A vertical line through (2,-3) is x=2.
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