Let $L$ be a line and $N$ a point in $L$ such that $\overrightarrow{ON}$ is a normal vector to $L$. Prove that an equation for $L$ is $$(\cos\theta)x+(\sin\theta)y=p$$ where $p=\|\overrightarrow{ON}\|$ and $\theta$ is the direction of $\overrightarrow{ON}.$
Try:
We know that if $\overrightarrow{N}=(a,b)$ is a normal vector to $L$, then an equation for $L$ is $ax+by=c$. So the only thing I have to prove is $\overrightarrow{ON}=(\cos\alpha,\sin\alpha)$ with $\alpha$ the direction of $\overrightarrow{ON}$, but how can I do that. Any help or it's already Of course $\overrightarrow{ON}$ can be written like this?
The vector $$(a,b):=\frac1p\,\overrightarrow{ON}=(\cos\theta,\sin\theta)$$ is also normal to $L$ hence (applying what you said you know) there exists a $c$ such that the equation of $L$ is $$ax+by=c.$$
Since moreover $N=(pa,pb)\in L,$ $$\begin{align}c&=a(pa)+b(pb)\\&=p\left(a^2+b^2\right)\\&=p,\end{align} $$ which ends the proof.
Alternatively (without applying what you said you know): a point $M=(x,y)$ belongs to $L$ iff $\overrightarrow{NM}\cdot\frac1p\overrightarrow{ON}=0,$ i.e. $$\begin{align}0&=\left(x-pa\right)a+\left(y-pb\right)b\\ &=ax+by-p\left(a^2+b^2\right)\\ &=ax+by-p. \end{align}$$