The problem asks us to find a line that is parallel, perpendicular, and neither parallel nor perpendicular to each of the given lines.

1. Problem Description

The problem asks us to find a line that is parallel, perpendicular, and neither parallel nor perpendicular to each of the given lines.

2. Solution Steps

Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of each other.

First, we rewrite all equations in the slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Line 11: $y = 3x + 4$. Slope $m_1 = 3$.

Parallel: $y = 3x + 5$

Perpendicular: $y = -\frac{1}{3}x + 2$

Neither: $y = x + 1$

Line 12: $2x - y = 8 \implies y = 2x - 8$. Slope $m_2 = 2$.

Parallel: $y = 2x + 1$

Perpendicular: $y = -\frac{1}{2}x + 3$

Neither: $y = x - 4$

Line 13: $3x + 4y + 12 = 0 \implies 4y = -3x - 12 \implies y = -\frac{3}{4}x - 3$. Slope $m_3 = -\frac{3}{4}$.

Parallel: $y = -\frac{3}{4}x + 2$

Perpendicular: $y = \frac{4}{3}x + 1$

Neither: $y = x + 5$

Line 14: $y = 3$. This is a horizontal line, with slope $m_4 = 0$.

Parallel: $y = 5$

Perpendicular: $x = 2$ (vertical line)

Neither: $y = x + 1$

3. Final Answer

Parallel to Line 11: $y = 3x + 5$

Perpendicular to Line 11: $y = -\frac{1}{3}x + 2$

Neither to Line 11: $y = x + 1$

Parallel to Line 12: $y = 2x + 1$

Perpendicular to Line 12: $y = -\frac{1}{2}x + 3$

Neither to Line 12: $y = x - 4$

Parallel to Line 13: $y = -\frac{3}{4}x + 2$

Perpendicular to Line 13: $y = \frac{4}{3}x + 1$

Neither to Line 13: $y = x + 5$

Parallel to Line 14: $y = 5$

Perpendicular to Line 14: $x = 2$

Neither to Line 14: $y = x + 1$