# Thread: Equation of a line derivation

1. ## 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

2. ## 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 $\boxed{b=-x_1 \tan(\alpha)+y_{1}}$

3. ## Re: Equation of a line derivation

Originally Posted by Ruun
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

4. ## 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.