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Thread: Proving quadrilateral is a rectangle algebraically

  1. #1
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    Proving quadrilateral is a rectangle algebraically

    How do you prove algebraically, that the following vertices are a rectangle. I can plot it on a graph, but I can't figure out how to prove using algebra.

    Help.
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  2. #2
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    Quote Originally Posted by GR9STUDENT View Post
    How do you prove algebraically, that the following vertices are a rectangle. I can plot it on a graph, but I can't figure out how to prove using algebra.

    Help.
    Would you happen to have the vertices in question? We typically look for equal lengths and equal slopes...
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  3. #3
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    Sorry, I did not list the vertices, they are (1,0) (-1,5) (4,7)(6,2)
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by GR9STUDENT View Post
    Sorry, I did not list the vertices, they are (1,0) (-1,5) (4,7)(6,2)
    1 way to do it:

    Let the given points correspond to $\displaystyle A,B,C,\text{ and }D$ respectively.

    Then to show that $\displaystyle ABCD$ is a rectangle, we must show that each interior angle measures $\displaystyle 90^{\circ}$ . To do this, we must need to only compare the slopes of line segments $\displaystyle AB$ with $\displaystyle BC$, and $\displaystyle BC$ with $\displaystyle CD$ and etc...

    Do you get what I'm saying?
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  5. #5
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    No, I am lost on this one.

    I understand that the rectangle has four 90 angles, and that the slope of each pair of lines will equal each other. But I don't know how to show with forumula.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by GR9STUDENT View Post
    No, I am lost on this one.

    I understand that the rectangle has four 90 angles, and that the slope of each pair of lines will equal each other. But I don't know how to show with forumula.
    OK.

    Perpindicular lines will have slopes that are the negative recirocal respective to each other.

    So $\displaystyle AB$ has slope $\displaystyle m_1=\frac{5-0}{-1-1}=-\frac{5}{2}$

    and $\displaystyle BC$ has slope $\displaystyle m_2=\frac{7-5}{4-(-1)}=\frac{2}{5}$

    Now, since $\displaystyle -\frac{1}{m_1}=m_2$ it follows that $\displaystyle AB$ is perpindicular to $\displaystyle BC$.
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  7. #7
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    Smile

    Thankyou, that makes sense.
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