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Math Help - Coordinate Geometry

  1. #1
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    Coordinate Geometry

    Three points have coordinates A(-2,-1), B(10,6) and C(a,-6). Given AB=BC find the possible values of a. If the figure ABCD is a rhoMbus, find the coordinates of D.
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    Quote Originally Posted by Ilsa View Post
    Three points have coordinates A(-2,-1), B(10,6) and C(a,-6). Given AB=BC find the possible values of a. If the figure ABCD is a rhoMbus, find the coordinates of D.
    AC= \sqrt{(-2-a)^2+ (-1+ 6)^2}= \sqrt{4+ 4a+ a^2+ 26}= \sqrt{a^2+ 4a+ 30}
    BC= \sqr{(10- a)^2+ (6+ 6)^2}= \sqrt{100- 20a+ a^2+ 144}= \sqrt{a^2- 20a+ 244}
    Since AC= BC, we must have a^2+ 4a+ 30}= a^2- 20a+ 244. Subtracting a^2- 20a+ 30 from both sides leaves 24a= 214.

    The vector BC= <a, -6>- <10, 6>= <a- 10-, -12> is parallel to the vector AD. Find the coordinates of D by add BC to the corodinates of A.
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