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Math Help - coordinate geometry-

  1. #1
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    Exclamation coordinate geometry-

    The points have the following coordinates:
    A(3,-2)
    B(p,3)
    C(6,2)
    D(q,r)
    The figure ABCD is a rhombus. Find the values of p, q and r.
    Even though I used the midpoint formula and the distance formula, I could not deduce the values of p and q.
    with the midpoint formula, i did get the value for r (-3).
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Ilsa View Post
    The points have the following coordinates:
    A(3,-2)
    B(p,3)
    C(6,2)
    D(q,r)
    The figure ABCD is a rhombus. Find the values of p, q and r.
    Even though I used the midpoint formula and the distance formula, I could not deduce the values of p and q.
    with the midpoint formula, i did get the value for r (-3).

    Hi Ilsa,

    First find the slope of AC. That would be \dfrac{4}{3}

    Since the diagonals of a rhombus are perpendicular, the slope of BD would be -\dfrac{3}{4}

    Next, find the midpoint of AC. That would be M\left(\dfrac{9}{2}, 0\right)

    Now, use points M\left(\dfrac{9}{2}, 0\right)\: \text{and}\: B(p, 3) and slope -\dfrac{3}{4} to find the missing x-coordinate p.

    \dfrac{0-3}{\frac{9}{2}-p}=-\dfrac{3}{4}

    This all boils down to \boxed{p=\dfrac{1}{2}}

    Since, M is also the midpoint of BD, we can easily find the coordinates of D this way:

    \dfrac{q+\frac{1}{2}}{2}=\dfrac{9}{2} and \dfrac{r+3}{2}=0

    Solving the above will yield \boxed{q=\dfrac{17}{2}} and \boxed{r=-3}

    The coordinates of the vertices of your rhombus are:

    3, -2)" alt="A\3, -2)" />

    B\left(\dfrac{1}{2}, 3\right)

    6, 2)" alt="C\6, 2)" />

    D\left(\dfrac{17}{2}, -3\right)

    You can verify that ABCD is a rhombus by using the distance formula:

    AB  = BC = CD = AD =\sqrt{31.25}



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