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Thread: coordinate geometry rhombus

  1. #1
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    coordinate geometry rhombus

    prove diagonals of rhombus are perpendicular by using coordinate geometry.

    it says to let the vertices be (0,0), (a,0), (a+b,0), (b,c)

    how do i use the pythagorean theorem to show this? a^2+b^2=c^2
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  2. #2
    Super Member wingless's Avatar
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    In the diagram, the vertices are (0,0), (a,0), (b,c), (a+b,c). All edges measure a. The diagonals are T1 and T2.

    As you see, there's a right triangle with edges a, b and c. From pythagorean theorem, we find $\displaystyle a^2 = b^2 + c^2$.

    If two lines are perpendicular, the product of their gradients is -1.

    $\displaystyle \text{Gradient of T1} = \frac{\Delta y}{\Delta x} = \frac{c}{a+b}$

    $\displaystyle \text{Gradient of T2} = \frac{\Delta y}{\Delta x} = \frac{-c}{a-b}$

    Product of them $\displaystyle = \frac{c}{a+b} \frac{-c}{a-b} = -\frac{c^2}{a^2-b^2}$

    Remember we found $\displaystyle a^2 = b^2 + c^2$ before. Then, $\displaystyle c^2 = a^2 - b^2$

    $\displaystyle -\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1$
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  3. #3
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    i understand everything up until this point:

    Remember we found before. Then,
    $\displaystyle
    -\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1
    $

    how did you get -c^2/c^2 ?
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by wingless View Post
    [IMG]
    In the diagram, the vertices are (0,0), (a,0), (b,c), (a+b,c). All edges measure a. The diagonals are T1 and T2.

    As you see, there's a right triangle with edges a, b and c. From pythagorean theorem, we find $\displaystyle \boxed{a^2 = b^2 + c^2}$.
    From, $\displaystyle a^2 = b^2 + c^2$, we get to $\displaystyle c^2 = a^2 - b^2$

    Then, in $\displaystyle -\frac{c^2}{a^2 - b^2}$, we just replaced $\displaystyle a^2 - b^2$ with $\displaystyle c^2$.
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  5. #5
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    thanks.
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