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Math Help - Proof of y-intercept

  1. #1
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    Proof of y-intercept

    Hi all, I'm working through some problems at the moment and I am a little stuck on an algebraic rearrangement of a coordinate geometry proof. The problem is as follows:

    We have a line passing through two points A: (x1, y1) and B: (x2, y2) (where x1 doesn't equal x2). Let P: (x, y) be any point on the line. We can define the equation of the line as:

    [1] ((y-y1)/(x-x1))=((y2-y1)/x2-x1))

    Putting this into the form y = mx +c gives

    [2] m= ((y2-y1)/(x2-x1)) ... I get this, but then it says that

    [3] c= (((x2y1)-(x1y2))/(x2-x1)) ... I don't understand how we arrive at this expression from rearranging [1], could someone please help?

    Thanks,
    Adam
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  2. #2
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    Re: Proof of y-intercept

    It's pretty much just algebra- specifically the "distributive law".

    You have \frac{y- y_1}{x- x_1}= \frac{y_2- y_1}{x_2- x_1}.

    Multiply both sides by x- x_1: y- y_1= \frac{y_2- y_1}{x_2- x_1}\left(x- x_1\right)

    Now "distribute" the fraction on the right over the " x" and " x_1"

    y- y_1= \frac{y_2- y1}{x_2- x_1}x- \frac{y_2- y_1}{x_2- x_1}x_1

    That first part, on the right, \frac{y_2- y_1}{x_2- x_1}x is the "mx" which you say you understand.

    Add y_1 to both sides
    y= \frac{y_2- y1}{x_2- x_1}x- \frac{y_2- y_1}{x_2- x_1}x_1+ y_1

    and add y_1 to that last fraction. That part alone, c, will be
    -\frac{y_2- y_1}{x_2- x_1}x_1+ y_1= \frac{-x_1(y_2- y_1)}{x_2- x_1}+ \frac{y_1(x_2- x_1}{x_2- x_1}
    = \frac{-x_1y_2+ x_1y_1+ x_2y_1- x_1y_1}{x_2- x_1}

    The " x_1y_1" terms cancel leaving
    \frac{x_2y_1- x_1y_2}{x_2- x_1}
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  3. #3
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    Re: Proof of y-intercept

    That's brilliant, thanks for such a quick response. I was getting stuck at the "distribute" part. Thanks again
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