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Math Help - line segment

  1. #1
    Senior Member sfspitfire23's Avatar
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    line segment

    Given two end points of a line segment as (5, t) and (-5, -t). If the length of the line segment is 2sqrt(34), find the slope of the line?
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Given two end points of a line segment as (5, t) and (-5, -t). If the length of the line segment is 2sqrt(34), find the slope of the line?
    The point goes through the origin, and you can simplify by saying the length of the segment from (0,0) to (5,t) is sqrt(34). Apply Pythagorean theorem to find t, then slope will be t/5.

    Edit: There is not enough information. We do not know if t is positive or negative. So we get two possible answers, and can't know which one is true. (I guess just use the \pm sign.)
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  3. #3
    Senior Member sfspitfire23's Avatar
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    can I also write the equation of the line m=(y-y1)/(x-x1) using a point and the slope and then plug in the other point for values x and y to solve for t?
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    can I also write the equation of the line m=(y-y1)/(x-x1) using a point and the slope and then plug in the other point for values x and y to solve for t?
    And where are you using the information of the length of the line segment?
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  5. #5
    Senior Member sfspitfire23's Avatar
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    i wasn't ;-)
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    i wasn't ;-)
    You can't solve the problem without using that information..

    Also please note the edit in my first reply of the thread, regarding plus/minus.
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  7. #7
    Senior Member sfspitfire23's Avatar
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    i get plus/minus sqrt(111)/5
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    i get plus/minus sqrt(111)/5
    5^2 + t^2 = (sqrt(34))^2

    25 + t^2 = 34

    t^2 = 9

    t = \pm 3

    slope m = \pm \frac{3}{5}
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