Academy

Normal Form

p > 0 is the perpendicular distance from the origin to the line. α is the angle the perpendicular makes with the positive x-axis.
Derivation

Let p>0p > 0 be the perpendicular distance from the origin to the line, and α\alpha the angle this perpendicular makes with the positive xx-axis. The foot of the perpendicular is N(pcos⁡α,psin⁡α)N(p\cos\alpha, p\sin\alpha).

The line is perpendicular to ONON, so its slope is −cot⁡α-\cot\alpha. Using point-slope form at NN:

y−psin⁡α=−cos⁡αsin⁡α(x−pcos⁡α)y - p\sin\alpha = -\frac{\cos\alpha}{\sin\alpha}(x - p\cos\alpha)

Multiply by sin⁡α\sin\alpha:

ysin⁡α−psin⁡2α=−xcos⁡α+pcos⁡2αy\sin\alpha - p\sin^2\alpha = -x\cos\alpha + p\cos^2\alpha xcos⁡α+ysin⁡α=p(cos⁡2α+sin⁡2α)=px\cos\alpha + y\sin\alpha = p(\cos^2\alpha + \sin^2\alpha) = p xcos⁡α+ysin⁡α=p\boxed{x\cos\alpha + y\sin\alpha = p}

Conversion from general form

Given ax+by+c=0ax + by + c = 0, divide by a2+b2\sqrt{a^2+b^2}:

aa2+b2x+ba2+b2y=−ca2+b2\frac{a}{\sqrt{a^2+b^2}}x + \frac{b}{\sqrt{a^2+b^2}}y = \frac{-c}{\sqrt{a^2+b^2}}

Choose the sign so the right side is positive (p>0p > 0). Then:

p=∣c∣a2+b2p = \frac{|c|}{\sqrt{a^2+b^2}}

This immediately gives the perpendicular distance from the origin to any line ax+by+c=0ax + by + c = 0.

Why p>0p > 0 is required

The normal form uses polar coordinates to locate the foot of the perpendicular: α\alpha carries all directional information, sweeping the full 360°360°. Since α\alpha alone determines which direction the normal points — including "backwards" — pp has no directional duty left. It is a pure magnitude. Allowing p<0p < 0 would make the representation non-unique: the same line would be described by (p,α)(p, \alpha) and (−p,α+180°)(-p, \alpha + 180°). The constraint p>0p > 0 ensures every line has exactly one normal form.

Why divide by a2+b2\sqrt{a^2+b^2}

The general equation ax+by+c=0ax + by + c = 0 and the normal form xcos⁡α+ysin⁡α−p=0x\cos\alpha + y\sin\alpha - p = 0 describe the same line, so their coefficients are proportional — not necessarily equal. The proportionality constant kk satisfies:

cos⁡αa=sin⁡αb=−pc=k\frac{\cos\alpha}{a} = \frac{\sin\alpha}{b} = \frac{-p}{c} = k

From cos⁡α=ak\cos\alpha = ak and sin⁡α=bk\sin\alpha = bk, apply cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1:

k2(a2+b2)=1  ⟹  k=±1a2+b2k^2(a^2 + b^2) = 1 \implies k = \pm\frac{1}{\sqrt{a^2+b^2}}

Dividing by a2+b2\sqrt{a^2+b^2} is not an algebraic trick — it is the act of normalising the vector (a,b)(a, b) to unit length, forcing the coefficients of xx and yy to become direction cosines (cos⁡α,sin⁡α)(\cos\alpha, \sin\alpha). The constant term on the right then becomes pp automatically.

Note
The general equation $ax + by + c = 0$ and $6ax + 6by + 6c = 0$ define the same line — coefficients define a line by their ratios, not their absolute values. This is why equality between the coefficients of the two forms is impossible in general (e.g. $\cos\alpha = 3$ is impossible), while proportionality always holds.

Why (a,b)(a, b) is normal to the line

Take two distinct points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) on the line ax+by+c=0ax + by + c = 0. Both satisfy the equation:

ax1+by1+c=0,ax2+by2+c=0ax_1 + by_1 + c = 0, \qquad ax_2 + by_2 + c = 0

Subtracting:

a(x2−x1)+b(y2−y1)=0a(x_2 - x_1) + b(y_2 - y_1) = 0

This is the dot product (a,b)⋅(x2−x1, y2−y1)=0(a, b) \cdot (x_2 - x_1,\, y_2 - y_1) = 0. The vector (x2−x1,y2−y1)(x_2 - x_1, y_2 - y_1) points along the line. A zero dot product means (a,b)(a, b) is perpendicular to it — i.e. (a,b)(a, b) is normal to the line.

The general equation ax+by=−cax + by = -c is therefore a geometric statement: project any point (x,y)(x, y) onto the normal vector (a,b)(a, b), and the result is always the same constant. Normalising converts that constant into the actual perpendicular distance from the origin.