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Thread: Analytic Geometry Q5

  1. #1
    Member looi76's Avatar
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    Analytic Geometry Q5

    Question:
    Show the points $\displaystyle A(-1,0)$ , $\displaystyle B(5,2)$ , $\displaystyle C(8,7)$ & $\displaystyle D(2,5)$ are vertices of a parallelogram.

    I don't what I'm suppose to find the prove that it's a parallelogram. Should I find the slope m of the lines? or should I find the distance between the points? and what does vertices mean?!

    EDIT: This question is related to the lesson of Inclination and Slope of a line.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Yes, you have to find the slope of the lines and see that

    $\displaystyle m_{AB}=m_{CD}$ and $\displaystyle m_{BC}=m_{AD}$
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  3. #3
    Member looi76's Avatar
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    $\displaystyle A(-1,0)$ , $\displaystyle B(5,2)$ , $\displaystyle C(8,7)$ & $\displaystyle D(2,5)$

    Slope of line $\displaystyle AB$:

    $\displaystyle A(-1,0)$ & $\displaystyle B(5,2)$

    $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2-0}{5-(-1)} = \frac{1}{3}$


    Slope of line $\displaystyle CD$:

    $\displaystyle C(8,7)$ & $\displaystyle D(2,5)$

    $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5-7}{2-8} = \frac{1}{3}$


    Slope of line $\displaystyle BC$:

    $\displaystyle B(5,2)$ & $\displaystyle C(8,7)$

    $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7-2}{8-5} = \frac{5}{3}$


    Slope of line $\displaystyle AD$:

    $\displaystyle A(-1,0)$ & $\displaystyle D(2,5)$

    $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5-0}{2-(-1)} = \frac{5}{3}$


    So, this is the answer?
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  4. #4
    MHF Contributor red_dog's Avatar
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    If $\displaystyle m_{AB}=m_{CD}\Rightarrow AB\parallel CD$

    and if $\displaystyle m_{BC}=m_{AD}\Rightarrow BC\parallel AD$

    So ABCD is a parallelogram.
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