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Math Help - equation of line

  1. #1
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    equation of line

    Can someone help me with this problem:
    Find the equation of the line which passes through the points (1
    ,3)
    and (
    1, 1). Sketch the graph of the line.

    Please go step by step. Thank You.

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jea9 View Post
    Can someone help me with this problem:
    Find the equation of the line which passes through the points (1
    ,3)
    and (
    1, 1). Sketch the graph of the line.

    Please go step by step. Thank You.

    the equation of a line is of the form y = mx + b, where m is the slope (or gradient) and b is the y-intercept.

    if m is not given, we can find it if two points are given in the following way.

    let the points given be (x1,y1) and (x2,y2), then
    m = (y2 - y1)/(x2 - x1)

    after finding the value for m, we can use it and any one set of points to find b, by either plugging in all the values into y = mx + b (at this time, b will be the only thing unknown) or plugging the values into the point-slope form and solving for y.

    point-slope form: y - y1 = m(x - x1)

    so now let's do your problem.

    the points given are (1,-3) and (-1,1)

    => m = (y2 - y1)/(x2 - x1) = (1 - (-3))/(-1 - 1) = 4/-2 = -2

    using (x1,y1) = (-1,1), we have by the point-slope form:

    y - y1 = m(x - x1)
    => y - 1 = -2(x - (-1)) = -2x - 2
    => y = -2x - 1
    Last edited by Jhevon; May 8th 2007 at 09:42 AM.
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  3. #3
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    using (x1,y1) = (-1,1), we have by the point-slope form:

    y - y1 = m(x - x1)
    => y - 1 = -2(x - (-1)) = -2x - 2
    => y = -2x - 1[/quote]


    Wait a minute... i didnt understand this part fully... if its y-y1=m(x-x1) then wouldnt the minuses become pluses? and on this part "y - 1 = -2(x - (-1)) = -2x - 2" how did you get the -2 on the end?
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  4. #4
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    Re:

    RE:

    y = -2x - 1

    Sketch
    Attached Thumbnails Attached Thumbnails equation of line-graph-1.png  
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