# Normal Form

Here, we will discuss about the normal form or perpendicular form of the equation of straight lines.

Let a straight line intersect x-axis and y-axis at $A$ and $B$ respectively. Draw a perpendicular $ON$ on the line $AB$ from the origin which is of length $p$, i.e., $ON=p$. Let the angle $NOA$ made by the perpendicular $ON$ with the positive x-axis be $\alpha$.

$\angle OBN=90°-\angle BON=90°-(90°-\alpha)=\alpha$ $\text{From }\Delta OAN, \cos\alpha=\frac{ON}{OA}=\frac{p}{OA}$ $\therefore OA=\frac{p}{\cos\alpha}=\text{x-intercept}$ $\text{From }\Delta OBN, \sin\alpha=\frac{ON}{OB}=\frac{p}{OB}$ $\therefore OB=\frac{p}{\cos\alpha}=\text{y-intercept}$

Now, by using the double intercept form, the equation of straight line is, $\frac{x}{OA}+\frac{y}{OB}=1$ $\frac{x}{\frac{p}{\cos\alpha}}+\frac{y}{\frac{p}{\sin\alpha}}=1$ $\frac{x\cos\alpha}{p}+\frac{y\sin\alpha}{p}=1$ $\therefore x\cos\alpha+y\sin\alpha=p$ This form of the equation of a straight line is known as normal form or perpendicular form.

### Obtain the equation of the straight line through the point $\left(\frac{1}{\sqrt{3}},1\right)$ whose perpendicular distance from the origin is unity.

Let the equation of the line be $x\cos\alpha+y\sin\alpha=p$

Since it passes through the point $\left(\frac{1}{\sqrt{3}},1\right)$ and its perpendicular distance from the origin i.e. $p=1$, so by substitution, we have, $\frac{1}{\sqrt{3}}.\cos\alpha+1.\sin\alpha=1$ $\frac{1}{2}\cos\alpha+\frac{\sqrt{3}}{2}\sin\alpha=\frac{\sqrt{3}}{2}$ $\cos 60°\cos\alpha+\sin 60°\sin\alpha=\frac{\sqrt{3}}{2}$ $\cos(\alpha-60°)=\cos(\pm 30°)$ $\alpha-60°=\pm 30°$ $\alpha=60°\pm 30°$ $\therefore\alpha=90°,30°$

When $\alpha=90°$, the equation of the line is $x\cos 90°+y\sin 90°=1$ $\therefore y=1$

When $\alpha=30°$, the equation of the line is $x\cos 30°+y\sin 30°=1$ $x\frac{\sqrt{3}}{2}+y\frac{1}{2}=1$ $\therefore\sqrt{3}\text{ }x+y=2$