Every general linear equation A x + B y + C = 0 Ax+By+C=0 A x + B y + C = 0 (with A , B A,B A , B not both zero) can be rewritten in any of the other named forms, by comparing coefficients:
Slope-intercept form (needs B ≠ 0 B\ne0 B = 0 ): dividing by B B B , y = − A B x − C B y=-\dfrac{A}{B}x-\dfrac{C}{B} y = − B A x − B C , so
slope = − A B , y -intercept = − C B . \text{slope}=-\frac{A}{B}, \qquad y\text{-intercept}=-\frac{C}{B}. slope = − B A , y -intercept = − B C .
Intercept form (needs A , B , C A,B,C A , B , C all nonzero): rewriting as x − C / A + y − C / B = 1 \dfrac{x}{-C/A}+\dfrac{y}{-C/B}=1 − C / A x + − C / B y = 1 gives
x -intercept = − C A , y -intercept = − C B . x\text{-intercept}=-\frac{C}{A}, \qquad y\text{-intercept}=-\frac{C}{B}. x -intercept = − A C , y -intercept = − B C .
Normal form (needs A , B ≠ 0 A,B\ne0 A , B = 0 ): comparing A x + B y + C = 0 Ax+By+C=0 A x + B y + C = 0 with x cos α + y sin α = p x\cos\alpha+y\sin\alpha=p x cos α + y sin α = p (i.e. x cos α + y sin α − p = 0 x\cos\alpha+y\sin\alpha-p=0 x cos α + y sin α − p = 0 ), the coefficients must be proportional:
cos α A = sin α B = − p C = ± cos 2 α + sin 2 α A 2 + B 2 = ± 1 A 2 + B 2 . \frac{\cos\alpha}{A}=\frac{\sin\alpha}{B}=\frac{-p}{C}=\pm\frac{\sqrt{\cos^2\alpha+\sin^2\alpha}}{\sqrt{A^2+B^2}}=\pm\frac{1}{\sqrt{A^2+B^2}}. A c o s α = B s i n α = C − p = ± A 2 + B 2 c o s 2 α + s i n 2 α = ± A 2 + B 2 1 .
The sign (+ + + or − - − ) is chosen so that the resulting p p p comes out positive (since a normal length is always taken positive); with that choice,
cos α = ∓ A A 2 + B 2 , sin α = ∓ B A 2 + B 2 , p = ∣ C ∣ A 2 + B 2 . \cos\alpha=\frac{\mp A}{\sqrt{A^2+B^2}}, \qquad \sin\alpha=\frac{\mp B}{\sqrt{A^2+B^2}}, \qquad p=\frac{|C|}{\sqrt{A^2+B^2}}. cos α = A 2 + B 2 ∓ A , sin α = A 2 + B 2 ∓ B , p = A 2 + B 2 ∣ C ∣ . …