Results 1 to 2 of 2

Thread: coordinate geometry-

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    66

    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).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,550
    Thanks
    15
    Awards
    1
    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 $\displaystyle \dfrac{4}{3}$

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

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

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

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

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

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

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

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

    The coordinates of the vertices of your rhombus are:

    $\displaystyle A\3, -2)$

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

    $\displaystyle C\6, 2)$

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

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

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



    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. coordinate geometry
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Mar 11th 2011, 05:44 AM
  2. Coordinate geometry
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Apr 1st 2008, 11:40 AM
  3. Coordinate geometry
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Apr 1st 2008, 11:07 AM
  4. Coordinate geometry
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Apr 1st 2008, 07:58 AM
  5. coordinate geometry
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 30th 2008, 12:01 AM

Search Tags


/mathhelpforum @mathhelpforum