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Coordinate geometry

  1. Jul 24, 2004 #1
    just wanna ask a simple question

    a quadrilateral has the vertices A ( 1,4 ), B (9,5 ) , C ( 5,-2) and D (-3,-3). show that the quadrilateral is a rhombus

    what i do is find gradient AB, CD, AD and BC.
    and then state which and which have the same gradient... and hence the opposite sides are parallel.

    is that all i should do?
     
  2. jcsd
  3. Jul 24, 2004 #2

    AKG

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    Unless they're doing something tricky, the sides should be AB, BC, DC, and AD. Show that AB = DC, BC = AD, and |AB| = |BC|, and you're done. Basically a rhombus as two defining properties:

    1) all four sides have the same length
    2) opposite sides are parallel

    In truth, it's sufficient to simply show that all four sides have the same length, and property 2 will follow, so that's one alternative approach. I would say try both, and see which one feels easier/more efficient. As for your approach, I don't know what you mean by gradient, so I can't say whether it will work or not.
     
  4. Jul 24, 2004 #3
    The two sufficient conditions to prove are:

    1. All 4 sides are equal (opposite sides are parallel).
    2. The diagonals are not equal.

    AKG: By gradient, denian means slope of a line joining [tex](x_{1}, y_{1})[/tex] and [tex](x_{2}, y_{2})[/tex] defined as:

    [tex]
    m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}
    [/tex]

    Cheers
    Vivek
     
  5. Jul 26, 2004 #4
    thank you!
     
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