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Math Help - Analytic Geometry Q6

  1. #1
    Member looi76's Avatar
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    Analytic Geometry Q6

    Question:
    Using slopes, determine if the points lie on a straight line.
    a) (-1,-4) , (3,8) & (0,-1)
    b) (0,5) , (-1,3) & (-3,2)

    Attempt:

    a) (-1,-4) & (3,8)

    m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8-(-4)}{3-(-1)} = \frac{12}{4} = 3

    (3,8) & (0,-1)

    m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1-8}{0-3} = \frac{-9}{-3} = 3

    It is a straight line as the slopes are the same.



    b) (0,5) & (-1,3)

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

    (-3,2) & (-1,3)

    m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-2}{(-1)-3} = \frac{1}{-4} = -\frac{1}{4}

    It is not a straight line as the slopes don't match.



    are my answers correct?
    Last edited by looi76; January 31st 2009 at 03:31 AM. Reason: Correction for question
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  2. #2
    MHF Contributor
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    a) you data is (0,1) but you made the calculation with (0,-1)

    b) your calculation of the second slope is not correct
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  3. #3
    Member looi76's Avatar
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    a) I have corrected the question. Is my answer right?



    b) (-3,2) & (-1,3)

    \color{red}{m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-2}{(-1)-3} = \frac{1}{-4} = -\frac{1}{4}}

    m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2-3}{-3-1} = \frac{1}{4}

    It is still not a straight line, am I right?
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  4. #4
    MHF Contributor
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    a) OK

    b) The slope is still not correct ...
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