Normal Form

5.3.5 Normal Form

Goal. Let $L$ be a line, and let $ON$ be the perpendicular ("normal") dropped from the origin onto $L$, with $ON=p$ and ray $ON$ making angle $\alpha$ with the positive X-axis. Find the equation of $L$.

Proof. The foot of the normal, $N$, has coordinates $(p\cos\alpha, p\sin\alpha)$ (standard polar-to-Cartesian conversion, since $ON=p$ at angle $\alpha$). The slope of $ON$ is $\dfrac{p\sin\alpha - 0}{p\cos\alpha - 0} = \tan\alpha$. Since $L \perp ON$, the slope of $L$ is $-\cot\alpha = -\dfrac{\cos\alpha}{\sin\alpha}$, and $L$ passes through $N(p\cos\alpha,p\sin\alpha)$. By the point-slope form,

Multiplying through by $\sin\alpha$ and simplifying using $\sin^2\alpha+\cos^2\alpha=1$:

So the normal form is

Worked Example 1. The perpendicular from the origin to a line has length $5$ and makes an angle $30^\circ$ with the positive X-axis; find the line. Here $p=5$, $\alpha=30^\circ$: $x\cos30^\circ+y\sin30^\circ=5 \Rightarrow \dfrac{\sqrt3}{2}x+\dfrac12 y=5 \Rightarrow \sqrt3x+y-10=0.$

Worked Example 2. Reduce $\sqrt3x-y-2=0$ to normal form and find $p,\alpha$. Comparing with $ax+by+c=0$: $a=\sqrt3,b=-1,c=-2$, so $\sqrt{a^2+b^2}=\sqrt{3+1}=2$. Dividing the equation by $2$: $\dfrac{\sqrt3}{2}x - \dfrac12 y = 1$. Since $\cos330^\circ = \dfrac{\sqrt3}{2}$ and $\sin330^\circ = -\dfrac12$, this is $x\cos330^\circ+y\sin330^\circ=1$, so $p=1$, $\alpha=330^\circ$.

Worked Example 3. Find the equation of the line in each case:

1. Parallel to the X-axis, 3 units below it: lines parallel to the X-axis have the form $y=k$; being below the axis makes $k$ negative, so $y=-3$.
2. Through the origin with inclination $30°$: slope $=\tan30°=\dfrac{1}{\sqrt3}$, so $y=\dfrac{x}{\sqrt3}$, i.e. $x-\sqrt3y=0$.
3. Through $A(5,2)$ with slope $-6$: by point-slope form, $y-2=-6(x-5) \Rightarrow 6x+y-32=0$.
4. Through $A(2,-1)$ and $B(5,1)$: by the two-points form, $\dfrac{x-2}{5-2}=\dfrac{y+1}{1+1} \Rightarrow \dfrac{x-2}{3}=\dfrac{y+1}{2} \Rightarrow 2(x-2)=3(y+1) \Rightarrow 2x-3y-7=0$. …