The problem asks us to find the equations of lines given certain conditions. First, we need to find the equation of a line given its distance from the origin and the angle the normal line makes with the x-axis (problems 1-4). Then, we are given the normal form of the line and need to write it in general form (problems 5-8).

1. Problem Description

The problem asks us to find the equations of lines given certain conditions. First, we need to find the equation of a line given its distance from the origin and the angle the normal line makes with the x-axis (problems 1-4). Then, we are given the normal form of the line and need to write it in general form (problems 5-8).

2. Solution Steps

Problem 1:

Distance from origin $p = 4$, angle $w = 60^{\circ}$.

The normal form of the line is given by:

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

Substituting the values:

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

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

Multiplying by 2, we get:

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

Problem 2:

Distance from origin $p = 6$, angle $w = \frac{5\pi}{6}$.

The normal form of the line is given by:

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

Substituting the values:

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

$x (-\frac{\sqrt{3}}{2}) + y (\frac{1}{2}) - 6 = 0$

Multiplying by 2, we get:

$-\sqrt{3}x + y - 12 = 0$

or

$\sqrt{3}x - y + 12 = 0$

Problem 3:

Distance from origin $p = 2$, angle $w = 30^{\circ}$.

The normal form of the line is given by:

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

Substituting the values:

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

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

Multiplying by 2, we get:

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

Problem 4:

Distance from origin $p = 5$, angle $w = \frac{7\pi}{6}$.

The normal form of the line is given by:

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

Substituting the values:

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

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

Multiplying by 2, we get:

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

or

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

Problem 5:

$x \cos(45^{\circ}) - y \sin(45^{\circ}) + 2 = 0$

Substituting the values:

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

Multiplying by $\frac{2}{\sqrt{2}}$, we get:

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

Problem 6:

$x \cos(225^{\circ}) + y \sin(225^{\circ}) - \sqrt{2} = 0$

Substituting the values:

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

Multiplying by $-\frac{2}{\sqrt{2}}$, we get:

$x + y + 2 = 0$

Problem 7:

$x \cos(\frac{2\pi}{3}) - y \sin(\frac{2\pi}{3}) - 3 = 0$

Substituting the values:

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

Multiplying by -2, we get:

$x + \sqrt{3}y + 6 = 0$

Problem 8:

$x \cos(\frac{5\pi}{4}) - y \sin(\frac{5\pi}{4}) + 1 = 0$

Substituting the values:

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

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

Multiplying by $-\frac{2}{\sqrt{2}}$, we get:

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

3. Final Answer

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

2) $\sqrt{3}x - y + 12 = 0$

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

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

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

6) $x + y + 2 = 0$

7) $x + \sqrt{3}y + 6 = 0$

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