We are asked to find the equations of lines given their distance from the origin, $p$, and the angle, $w$, between the positive x-axis and the perpendicular line segment from the origin to the line. We are given the following problems: 1) Distance from the origin is 4 units and the angle is 60 degrees. 2) Distance from the origin is 6 units and the angle is $\frac{5\pi}{4}$. 3) $p = 2$ and $w = 30^{\circ}$. 4) $p = 5$ and $w = \frac{7\pi}{6}$.

1. Problem Description

We are asked to find the equations of lines given their distance from the origin, $p$, and the angle, $w$, between the positive x-axis and the perpendicular line segment from the origin to the line. We are given the following problems:

1) Distance from the origin is 4 units and the angle is 60 degrees.

2) Distance from the origin is 6 units and the angle is $\frac{5\pi}{4}$.

3) $p = 2$ and $w = 30^{\circ}$.

4) $p = 5$ and $w = \frac{7\pi}{6}$.

2. Solution Steps

The general form of the equation of a line, given its distance $p$ from the origin and the angle $w$ between the positive x-axis and the perpendicular line segment from the origin to the line, is:

$x \cos(w) + y \sin(w) - p = 0$

1) $p = 4$ and $w = 60^{\circ} = \frac{\pi}{3}$ radians. Then $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$. Substituting into the equation gives:

$x \cos(60^{\circ}) + y \sin(60^{\circ}) - 4 = 0$

$x \frac{1}{2} + y \frac{\sqrt{3}}{2} - 4 = 0$

$x + y\sqrt{3} - 8 = 0$

2) $p = 6$ and $w = \frac{5\pi}{4}$. Then $\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$ and $\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$. Substituting into the equation gives:

$x \cos(\frac{5\pi}{4}) + y \sin(\frac{5\pi}{4}) - 6 = 0$

$x(-\frac{\sqrt{2}}{2}) + y(-\frac{\sqrt{2}}{2}) - 6 = 0$

$-\sqrt{2}x - \sqrt{2}y - 12 = 0$

$\sqrt{2}x + \sqrt{2}y + 12 = 0$

3) $p = 2$ and $w = 30^{\circ} = \frac{\pi}{6}$ radians. Then $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$. Substituting into the equation gives:

$x \cos(30^{\circ}) + y \sin(30^{\circ}) - 2 = 0$

$x \frac{\sqrt{3}}{2} + y \frac{1}{2} - 2 = 0$

$\sqrt{3}x + y - 4 = 0$

4) $p = 5$ and $w = \frac{7\pi}{6}$. Then $\cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2}$ and $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$. Substituting into the equation gives:

$x \cos(\frac{7\pi}{6}) + y \sin(\frac{7\pi}{6}) - 5 = 0$

$x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) - 5 = 0$

$-\sqrt{3}x - y - 10 = 0$

$\sqrt{3}x + y + 10 = 0$

3. Final Answer

1) $x + y\sqrt{3} - 8 = 0$

2) $\sqrt{2}x + \sqrt{2}y + 12 = 0$

3) $\sqrt{3}x + y - 4 = 0$

4) $\sqrt{3}x + y + 10 = 0$