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Math Help - Proving a quadrilateral is a rectangle?

  1. #1
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    Proving a quadrilateral is a rectangle?

    I'm having a little trouble with a question I'm doing. I'll write it below.

    The vertices of a quadrilateral are C(-2,4), D(3,9), E(6,6), and F(1,1).

    a) Show that the quadrilateral is a rectangle.

    First I calculated the length of its four sides.

    CD = square root of 50
    DE = square root of 18
    EF = square root of 50
    CF = square root of 18

    Ok, so far so good. Now I have to show that it has a right angle, which would require two adjacent sides being perpendicular.

    mCD = 1, but mDE = -6/0?

    Did I mess up somewhere? Any help/advice is appreciated.
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  2. #2
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    Quote Originally Posted by mathdonkey View Post
    mCD = 1, but mDE = -6/0?
    Now, what are you doing here?


    If you are familiair with vectors, try to express each side as a vector. For an example, \bar{CD}=\begin{pmatrix}3-(-2) \\ 9-4\end{pmatrix}=\begin{pmatrix}5 \\ 5\end{pmatrix}
    and \bar{DE}=\begin{pmatrix}6-3 \\ 6-9\end{pmatrix}=\begin{pmatrix}3 \\ -3\end{pmatrix}

    For \bar{CD} and \bar{DE} to perpendicular we must have that the scalar product is zero.

    \bar{CD} \cdot \bar{DE}=\begin{pmatrix}5 \\ 5\end{pmatrix} \cdot \begin{pmatrix}3 \\ -3\end{pmatrix}= 5(3)+5(-3)=0, so the vectors are perpendicular.
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  3. #3
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    Quote Originally Posted by Aliquantus View Post
    Now, what are you doing here?


    If you are familiair with vectors, try to express each side as a vector. For an example, \bar{CD}=\begin{pmatrix}3-(-2) \\ 9-4\end{pmatrix}=\begin{pmatrix}5 \\ 5\end{pmatrix}
    and \bar{DE}=\begin{pmatrix}6-3 \\ 6-9\end{pmatrix}=\begin{pmatrix}3 \\ -3\end{pmatrix}

    For \bar{CD} and \bar{DE} to perpendicular we must have that the scalar product is zero.

    \bar{CD} \cdot \bar{DE}=\begin{pmatrix}5 \\ 5\end{pmatrix} \cdot \begin{pmatrix}3 \\ -3\end{pmatrix}= 5(3)+5(-3)=0, so the vectors are perpendicular.
    This is what I had,

    mCD=-2-4/3-9
    =-6/-6
    =1

    mDE=3-9/6-6
    =-6/0

    I think I'm overlooking something, I just can't figure out what.
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  4. #4
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    Hello, mathdonkey!

    Your reasoning is absolutely correct, but . . .


    m_{CD} \:= \:1\text{, but }m_{DE} \:= \:{\bf\frac{\text{-}6}{0}}\; ? . . . . How?

    We have: . D(3,9)\text{ and }E(6,6)

    Therefore: . m_{DE} \;=\;\frac{6-9}{6-3} \;=\;\frac{\text{-}3}{3} \;=\;-1

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, mathdonkey!

    Your reasoning is absolutely correct, but . . .



    We have: . D(3,9)\text{ and }E(6,6)

    Therefore: . m_{DE} \;=\;\frac{6-9}{6-3} \;=\;\frac{\text{-}3}{3} \;=\;-1

    Hi Soroban,

    Thanks for looking over my work!

    I see what you did to get an answer of -1, but the method you used is different than the one I used. In yours the vertice is seperated and combined with the other vertice, in mine it isn't. I don't understand why this is? Sorry if I'm not being clear enough, I don't talk math often.
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  6. #6
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    Quote Originally Posted by mathdonkey View Post
    I'm having a little trouble with a question I'm doing. I'll write it below.

    The vertices of a quadrilateral are C(-2,4), D(3,9), E(6,6), and F(1,1).

    ... two adjacent sides being perpendicular.

    mCD = 1, but mDE = -6/0?

    Did I mess up somewhere? Any help/advice is appreciated.
    Hello,

    if you have 2 points P_1\left(x_1,\ y_1 \right) and P_2\left(x_2,\ y_2 \right) then the slope of the line which connect these 2 points is calculated by:

    m_{P_1, P_2} = \frac{y_1 - y_2}{x_1 - x_2}. .......... That's the formula Soroban used in his post:

    m_{D, E}=\frac{9 - 6}{3 - 6} = \frac3{-3}= -1

    I can't imagine how you got a zero in the denominator
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  7. #7
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    Quote Originally Posted by earboth View Post
    Hello,

    if you have 2 points P_1\left(x_1,\ y_1 \right) and P_2\left(x_2,\ y_2 \right) then the slope of the line which connect these 2 points is calculated by:

    m_{P_1, P_2} = \frac{y_1 - y_2}{x_1 - x_2}. .......... That's the formula Soroban used in his post:

    m_{D, E}=\frac{9 - 6}{3 - 6} = \frac3{-3}= -1

    I can't imagine how you got a zero in the denominator
    I wasn't doing it right.

    Thanks for the help guys.
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  8. #8
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    Oops, ignore this post.

    Thanks again for the help guys!
    Last edited by mathdonkey; January 21st 2008 at 01:47 PM.
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  9. #9
    Senior Member JaneBennet's Avatar
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    Here’s something you can try. Multiply the points C, D, E, F by the matrix \left(\begin{array}{rr}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\\<br />
-\,\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right), which is a clockwise rotation of 45 about the origin.

    \frac{1}{\sqrt{2}}\left(\begin{array}{rr}1&1\\-1&1<br />
\end{array}\right)\left(\begin{array}{rrrr}-2&3&6&1\\4&9&6&1<br />
\end{array}\right)=\left(\begin{array}{rrrr}\sqrt{  2}&6\sqrt{2}&6\sqrt{2}&\sqrt{2}\\3\sqrt{2}&3\sqrt{  2}&0&0<br />
\end{array}\right)

    Hence we have rotated the vertices of the quadrilateral to \mathrm{C}'(\sqrt{2},3\sqrt{2}), \mathrm{D}'(6\sqrt{2},3\sqrt{2}), \mathrm{E}'(6\sqrt{2},0), \mathrm{F}'(\sqrt{2},0). Now it is much easier to see that \mathrm{C}'\mathrm{D}'\mathrm{E}'\mathrm{F}' is a rectangle.
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