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The equations $$Ax+By+C=0$$ and $$x\cos \alpha + y\sin \alpha - p=0$$ will represent one and the same straight line if their corresponding coefficients are proportional: $$\dfrac {\cos \alpha}{A} = \dfrac {\sin \alpha}{B}=\dfrac {-p}{C}=k(\textrm {say})$$ So that: $$\cos \alpha=Ak$$ $$\sin \alpha =Bk$$ $$-p=Ck$$ Now, $$A^2k^2+B^2k^2=1$$

$$k^2=\dfrac {1}{A^2+B^2}$$ $$k=\pm \dfrac {1}{\sqrt {A^2+B^2}}$$

If $C>0$, then $$k=-\dfrac {1}{\sqrt {A^2+B^2}}$$

Here I don't understand the portion after "If $C>0$, then……

pi-π
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2 Answers2

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By convention, $p\ge 0$, since it is the distance from the origin to the line. (The normal vector to the line is $(\cos\alpha,\sin\alpha)$, and $(x,y)\cdot(\cos\alpha,\sin\alpha) = x\cos\alpha+y\sin\alpha$ gives this distance.) So if $k=-p/C$ and $p$ and $C$ are positive, then $k$ must be negative.

Ted Shifrin
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In your derivation you get to the point $$ k=\pm \frac {1}{\sqrt{A^2+B^2}} $$ where two solutions are possible, distinguished by their sign, and you have to decide which sign is the valid one. The relation for $C$ was $$ -\frac{p}{C} = k $$ So if $p > 0$, then $C$ and $k$ have opposite signs, if $p < 0$ then $C$ and $k$ have the same sign.

mvw
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