Equation of a Straight Line in Normal Form

The equation of a straight line can be expressed in various forms, such as slope-intercept form, point-slope form, two-point form, and normal form. The normal form of a straight line is particularly useful in certain geometric applications, such as finding the shortest distance from a point to a line.

Normal Form of a Straight Line

The normal form of the equation of a straight line is given by:

[ x \cos \theta + y \sin \theta = p ]

where:

• $\theta$ (theta) is the angle made by the perpendicular from the origin to the line with the positive direction of the x-axis.
• $p$ is the length of the perpendicular from the origin to the line (always taken as positive).

Important Points

• The angle $\theta$ is measured in the anticlockwise direction from the positive x-axis to the line.
• The value of $\theta$ lies in the range $[0, 2\pi)$ or $[0^\circ, 360^\circ)$.
• The value of $p$ is the shortest distance from the origin to the line and is always non-negative.

Derivation of Normal Form

The normal form can be derived from the general equation of a straight line $Ax + By + C = 0$. By dividing the entire equation by $\sqrt{A^2 + B^2}$, we get:

[ \frac{A}{\sqrt{A^2 + B^2}}x + \frac{B}{\sqrt{A^2 + B^2}}y = -\frac{C}{\sqrt{A^2 + B^2}} ]

Let $\cos \theta = \frac{A}{\sqrt{A^2 + B^2}}$ and $\sin \theta = \frac{B}{\sqrt{A^2 + B^2}}$, and let $p = -\frac{C}{\sqrt{A^2 + B^2}}$. Then the equation becomes:

[ x \cos \theta + y \sin \theta = p ]

Differences and Important Points

Aspect	Normal Form	General Form
Equation	$x \cos \theta + y \sin \theta = p$	$Ax + By + C = 0$
Parameters	$\theta$ (angle), $p$ (perpendicular distance)	$A$, $B$, $C$ (coefficients)
Angle Range	$[0, 2\pi)$ or $[0^\circ, 360^\circ)$	Not applicable
Distance $p$	Always positive	Not applicable
Orientation	Defined by $\theta$	Defined by the ratio $\frac{-A}{B}$

Examples

Example 1: Find the Normal Form

Given the general equation of a line $3x + 4y - 12 = 0$, find its normal form.

Solution:

First, we find the values of $\cos \theta$ and $\sin \theta$:

[ \cos \theta = \frac{3}{\sqrt{3^2 + 4^2}} = \frac{3}{5} ] [ \sin \theta = \frac{4}{\sqrt{3^2 + 4^2}} = \frac{4}{5} ]

Next, we find the value of $p$:

[ p = -\frac{-12}{\sqrt{3^2 + 4^2}} = \frac{12}{5} ]

Thus, the normal form is:

[ x \left(\frac{3}{5}\right) + y \left(\frac{4}{5}\right) = \frac{12}{5} ]

Example 2: Find the Distance from the Origin

Given the line in normal form $x \cos 60^\circ + y \sin 60^\circ = 10$, find the shortest distance from the origin to the line.

Solution:

The shortest distance from the origin to the line is given by the value of $p$ in the normal form equation. Here, $p = 10$.

Therefore, the shortest distance from the origin to the line is 10 units.

Example 3: Convert to General Form

Convert the normal form $x \cos \frac{\pi}{4} + y \sin \frac{\pi}{4} = \sqrt{2}$ to the general form.

Solution:

We know that $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Substituting these values into the normal form equation, we get:

[ x \left(\frac{\sqrt{2}}{2}\right) + y \left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} ]

Multiplying through by $2$, we obtain the general form:

[ \sqrt{2}x + \sqrt{2}y - 2 = 0 ]

or

[ x + y - \sqrt{2} = 0 ]

In conclusion, the normal form of a straight line provides a clear geometric interpretation of the line's orientation and distance from the origin. It is a useful form for solving problems involving distances and angles related to straight lines.