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Math Help - Equation in slop intercept forum

  1. #1
    Junior Member Sammyj's Avatar
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    Equation in slop intercept forum

    I have the answer, but I am stuck in the middle of the problem

    Through (4,2) and (1,3)
    m= 3 - 2 / 1 - 4 = - 1/3

    y - 2 = (-1/3)x (x-4)

    y - 2 = this is where I get stuck. I know the answer is y = (-1/3)x + 11/3
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  2. #2
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    Quote Originally Posted by Sammyj View Post
    I have the answer, but I am stuck in the middle of the problem

    Through (4,2) and (1,3)
    m= 3 - 2 / 1 - 4 = - 1/3

    y - 2 = (-1/3)x (x-4)

    y - 2 = this is where I get stuck. I know the answer is y = (-1/3)x + 11/3
    I get a different answer.

    The general form is:

    y - y_1= m(x-x_1)

    Substituting values:

    y-2 = -\frac13 ( x - 4)

    Opening the bracket:

    y-2 = -\frac13x + \frac43

    Adding 2 to both sides:

    y = -\frac13 + \frac{10}{3}
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  3. #3
    Junior Member Sammyj's Avatar
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    Thanks
    Last edited by Sammyj; September 8th 2008 at 10:57 AM. Reason: error
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  4. #4
    Junior Member Sammyj's Avatar
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    How about through (1/2, 5/3) and (3, 1/16)

    (1/16) - (5/3) / (3) - (1/2)

    I understand how to do them until they are fractions. I know the answer is y = (2/3)x-2 but don't know how to work it out.
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    Quote Originally Posted by Sammyj View Post
    How about through (1/2, 5/3) and (3, 1/16)

    (1/16) - (5/3) / (3) - (1/2)

    I understand how to do them until they are fractions. I know the answer is y = (2/3)x-2 but don't know how to work it out.
    \Delta y = \frac{1}{16} - \frac{5}{3} = -\frac{77}{48}

    \Delta x = 3 - \frac12 = \frac52

    m = \frac{\Delta y}{\Delta x} = \frac{-\frac{77}{48}}{\frac52}

    m = -\frac{77}{48} \div {\frac52} = -\frac{77}{48} \times \frac{2}{5} = -\frac{77}{120}

    Your gradient is m = -\frac{77}{120}

    Use your normal apprach to find the equation. Substitute appropriate number into the general form, y - y_1 = m(x-x_1).

    EDIT: Sammyj, is your question posted correctly? The gradient is different to the answer that you posted.
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