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Math Help - Rumbus

  1. #1
    Junior Member
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    Smile Rumbus

    Hi there, I have a math problem that I am having bit of a problem with. The question said that the equation of two adjacent sides of a rhombus are y= 2x +4, y= -\frac{1}{3}+4. If (12,0) is one vertex and all vertices have positive coordinates find the coordinates of the other three vertices. Can someone please help me with this problem I am trying to get the answer in radix. thank you in advance.
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  2. #2
    Super Member

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    Hello, scrible!

    The equation of two adjacent sides of a rhombus are y\,=\, 2x +4,\;y\,=\,-\tfrac{1}{3}+4.
    If (12,0) is one vertex and all vertices have positive coordinates,
    find the coordinates of the other three vertices.
    We are given quite a lot of information . . .
    Code:
            |     C
            |     o
            |    *    *
            |   *         *
            |  *              o D
            | *              *
          B |*              *
       (0,4)o              *
            |   *         *
            |       *    *
        - - + - - - - - o - - - - - - -
            |        (12,0)
            |           A

    We are given vertex A(12,0).

    Both lines have a y-intercept of 4.
    . . So another vertex is: . \boxed{B(0,4)}

    Side AB has length: . \sqrt{(0-12)^2 + (4-0)^2} \:=\:\sqrt{160} \:=\:4\sqrt{10}

    Point C is on the line y \:=\:2x+4
    Point C has coordinates: . C(x,\:2x+4)


    Since side BC = side AB, we have: . \sqrt{(x-0)^2 + (2x+4-4)^2} \:=\:\sqrt{160}

    . . \sqrt{x^2 + 4x^2} \:=\:\sqrt{160}\quad\Rightarrow\quad \sqrt{5x^2} \:=\:\sqrt{160} \quad\Rightarrow\quad x^2 \:=\:32 \quad\Rightarrow\quad x \:=\:4\sqrt{2}

    Then: . y \:=\:2(4\sqrt{2}) + 4 \quad\Rightarrow\quad y \:=\:8\sqrt{2} + 4

    Therefore: . \boxed{C\left(4\sqrt{2},\;8\sqrt{2}+4\right) }



    Since AD \parallel BC\,\text{ and }\,AD \,=\, BC, we can use this trick.

    Going from B to C, we moved: . \text{right }4\sqrt{2}\:\text{ and }\,\text{ up }8\sqrt{2}

    Going from A(12,0) to D, we will make the same moves.

    . . And we arrive at: . \boxed{D\left(12+4\sqrt{2},\;8\sqrt{2}\right)}

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