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Math Help - Equation of a line derivation

  1. #1
    Senior Member bugatti79's Avatar
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    Equation of a line derivation

    Hi Folks,

    Just wondering how one derives the equation of a line y=mx+b from this general expression f(x,y)=a+bx+cy. I believe this expression represents plane geometry.

    Any thoughts
    Regards
    bugatti
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  2. #2
    Member Ruun's Avatar
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    Re: Equation of a line derivation

    Pick one point on the plane, say (x_{1},y_{1}) and other one (x,y). We will try to see the relation between x and y Now impose that they are in a line:


    Let \alpha be the angle between the black and the red segments. We have \tan(\alpha)=\frac{\Delta y}{\Delta x} where \Delta y=y-y_1 and \Delta x=x-x_1. Then

    \tan(\alpha)(x-x_1)=(y-y_1) \rightarrow y-y_1=x\tan(\alpha)}-x_{1}\tan (\alpha)

    This is

    \boxed{y=mx+b} where \boxed{m=\tan(\alpha)} and \boxed{b=-x_1 \tan(\alpha)+y_{1}}
    Last edited by Ruun; March 29th 2013 at 04:34 AM.
    Thanks from bugatti79
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Equation of a line derivation

    Quote Originally Posted by Ruun View Post
    Pick one point on the plane, say (x_{1},y_{1}) and other one (x,y). We will try to see the relation between x and y Now impose that they are in a line:


    Let \alpha be the angle between the black and the red segments. We have \tan(\alpha)=\frac{\Delta y}{\Delta x} where \Delta y=y-y_1 and \Delta x=x-x_1. Then

    \tan(\alpha)(x-x_1)=(y-y_1) \rightarrow y-y_1=x\tan(\alpha)}-x_{1}\tan (\alpha)

    This is

    \boxed{y=mx+b} where \boxed{m=\tan(\alpha)} and \boxed{b=-x_1 \tan(\alpha)+y_{1}}
    Hi Runn,

    Thanks for the reply. Does this line have any connection with f(x,y)=a+bx+cy? Can it be derived from this?

    Thanks
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  4. #4
    MHF Contributor

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    Re: Equation of a line derivation

    f(x,y)= z= ax+ by+ c is the equation of a plane in three dimensions. What do you mean by "derive the equation of a line"? Which line? There exist an infinite number of lines in that plane. We could, for example, get the equation of the line where it intersects the xy plane by setting z= 0: 0= ax+ by+ c which is the same as -by= ax+ c and then divide both sides by -b: y= (-a/b)x- (c/b) which is of the form "y= mx+b" with m= -a/b and "b"= -c/b.
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