Straight Lines - Normal Form of a Line

Find the equation of the line for which the length of the perpendicular from the origin is $5$ units and the angle which the perpendicular makes with the positive x-axis is $30^{\circ}$.

Step 1: Identify the given values. We have $p = 5$ and $\omega = 30^{\circ}$. Step 2: Use the Normal Form equation $x \cos \omega + y \sin \omega = p$. Step 3: Substitute the values: $x \cos 30^{\circ} + y \sin 30^{\circ} = 5$. Step 4: Evaluate trigonometric ratios: $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$. Step 5: Substitute these back into the equation: $x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 5$. Step 6: Simplify by multiplying the entire equation by $2$: $\sqrt{3}x + y = 10$.

This is a direct application of the normal form. We simply plug the distance $p$ and the angle $\omega$ into the standard equation and simplify.

Reduce the equation $\sqrt{3}x + y - 8 = 0$ into normal form. Find the values of $p$ and $\omega$.

Step 1: Rewrite the equation as $\sqrt{3}x + y = 8$. Here, the constant term on the right is already positive ($8 > 0$). Step 2: Calculate $\sqrt{A^2 + B^2}$, where $A = \sqrt{3}$ and $B = 1$. So, $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$. Step 3: Divide the entire equation by $2$: $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2}$, which simplifies to $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4$. Step 4: Compare with $x \cos \omega + y \sin \omega = p$. We find $p = 4$. Step 5: Determine $\omega$ from $\cos \omega = \frac{\sqrt{3}}{2}$ and $\sin \omega = \frac{1}{2}$. Both are positive, so $\omega$ is in the first quadrant. $\omega = 30^{\circ}$ or $\frac{\pi}{6}$.

To reduce a general equation to normal form, we divide by the magnitude of the coefficients' vector $\sqrt{A^2+B^2}$. This normalizes the coefficients so they represent the sine and cosine of the same angle.

Given $p = 4$ and $\omega = 150^{\circ}$. The normal form is $x \cos \omega + y \sin \omega = p$. Substituting the values: $x \cos 150^{\circ} + y \sin 150^{\circ} = 4$ Since $\cos 150^{\circ} = \cos(180^{\circ} - 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$ and $\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$, $x(-\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$ Multiplying by $2$: $-\sqrt{3}x + y = 8$ or $\sqrt{3}x - y + 8 = 0$.

We identify $p$ and $\omega$ from the question and substitute them into the standard normal form equation. Trigonometric values for $150^{\circ}$ are calculated using reference angles in the second quadrant.

The given equation is $x - y - 4 = 0$, which can be written as $x - y = 4$. Here $A = 1, B = -1, C = -4$. Divide both sides by $\sqrt{A^2 + B^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}}$ $x(\frac{1}{\sqrt{2}}) + y(-\frac{1}{\sqrt{2}}) = 2\sqrt{2}$ Comparing with $x \cos \omega + y \sin \omega = p$: $p = 2\sqrt{2}$ $\cos \omega = \frac{1}{\sqrt{2}}$ and $\sin \omega = -\frac{1}{\sqrt{2}}$ Since $\cos$ is positive and $\sin$ is negative, $\omega$ lies in the 4th quadrant. $\omega = 360^{\circ} - 45^{\circ} = 315^{\circ}$ (or $\frac{7\pi}{4}$ radians).

To reduce to normal form, divide by $\sqrt{A^2+B^2}$. Ensure the constant $p$ is positive. Determine the quadrant of $\omega$ based on the signs of $\sin$ and $\cos$.