Results 1 to 2 of 2

Thread: Rumbus

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    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 $\displaystyle y= 2x +4$, $\displaystyle y= -\frac{1}{3}+4$. 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, scrible!

    The equation of two adjacent sides of a rhombus are $\displaystyle y\,=\, 2x +4,\;y\,=\,-\tfrac{1}{3}+4$.
    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 $\displaystyle A(12,0).$

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

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

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


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

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

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

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



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

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

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

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

    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum