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Math Help - 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 a^2 = b^2 + c^2.

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

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

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

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

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

    -\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,
    <br />
-\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1<br />

    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 \boxed{a^2 = b^2 + c^2}.
    From, a^2 = b^2 + c^2, we get to c^2 = a^2 - b^2

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