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Math Help - Standard form of the equation a line

  1. #1
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    Standard form of the equation a line

    Alright, I know I've done this before but I can't remember how to set it up to save my life.

    Write in standard form the equation of line l who lies on points (2,5) and (7,8)

    Would appreciate an explanation on how to do this, thanks.
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  2. #2
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    Given two points and the slope between them, you can write the line in point slope form: y-y_1=m(x-x_1), where m is the slope. The slope is rise over run, which is \frac{8-5}{7-2}=\frac{3}{5}. Now, we choose any given point and plug it into the equation. Let's choose (2,5), and we get y-5=\frac{3}{5}(x-2). From there you can manipulate the equation to standard form: ax+by=c, where a, b, c are integers.
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  3. #3
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    Quote Originally Posted by MathTooHard View Post
    Given two points and the slope between them, you can write the line in point slope form: y-y_1=m(x-x_1), where m is the slope. The slope is rise over run, which is \frac{8-5}{7-2}=\frac{3}{5}. Now, we choose any given point and plug it into the equation. Let's choose (2,5), and we get y-5=\frac{3}{5}(x-2). From there you can manipulate the equation to standard form: ax+by=c, where a, b, c are integers.
    So just to make sure..
    <br />
y-5=\frac{3}{5}(x-2)<br />

    What do I do with the y-5 part?

    y-5=\frac{3}{5}x+1\frac{1}{5}

    That's where I'm at right now..
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  4. #4
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    Add 5 to both sides, by changing 5 into a number with a denominator of 5

    y-5+5=\frac{3}{5}x+\frac{6}{5}+5

    y=\frac{3}{5}x+\frac{6}{5}+\frac{25}{5}

    y=\frac{3}{5}x+\frac{31}{5}

    y=\frac{3}{5}x+6\frac{1}{5}

    EDIT: Woops, didn't bother checking, just used what he/she already had.
    Last edited by user_5; November 26th 2009 at 02:00 AM.
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  5. #5
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    Quote Originally Posted by user_5 View Post
    Add 5 to both sides, by changing 5 into a number with a denominator of 5

    y-5+5=\frac{3}{5}x+\frac{6}{5}+5

    y=\frac{3}{5}x+\frac{6}{5}+\frac{25}{5}

    y=\frac{3}{5}x+\frac{31}{5}

    y=\frac{3}{5}x+6\frac{1}{5}
    This is not correct.

    y-5 = \frac{3}{5}x-\frac{6}{5}

    y = \frac{3x-6}{5}+5

    y = \frac{3x-6}{5}+\frac{25}{5}

    y = \frac{3x-6+25}{5}

    y = \frac{3x+19}{5}

    In linear form y= ax + b:

    y = \frac{3}{5}x+\frac{19}{5}

    Plug your two example pairs (2,5) and (7,8) to verify this is correct.
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