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.

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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 \[x\cos 30°+y\sin 30°=1\] \[x\frac{\sqrt{3}}{2}+y\frac{1}{2}=1\] \[\therefore\sqrt{3}\text{ }x+y=2\]

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