To solve the problem, we need to identify lines that are either parallel or perpendicular to the given line $A$, which has the equation:

$y = 4x - 3$

1. Parallel Lines:

Lines are parallel if they have the same slope. The slope-intercept form of a line is $y = mx + b$, where $m$ is the slope. For line $A$, the slope is $m = 4$.

Thus, any line with slope $4$ is parallel to line $A$.

2. Perpendicular Lines:

Lines are perpendicular if the product of their slopes is $-1$. So, for a line to be perpendicular to line $A$, its slope $m'$ must satisfy:

$4 \times m' = -1 \quad \Rightarrow \quad m' = -\frac{1}{4}$

Thus, any line with slope $-\frac{1}{4}$ is perpendicular to line $A$.

Now, let's analyze the given equations:

1. $y = 4x + 2$ ⟶ Slope is $4$ (Parallel to line $A$).
2. $y = -\frac{1}{4}x + 5$ ⟶ Slope is $-\frac{1}{4}$ (Perpendicular to line $A$).
3. $y = 4x + 9$ ⟶ Slope is $4$ (Parallel to line $A$).
4. $y + 4 = -\frac{1}{4}(x - 3)$ ⟶ This can be rewritten as $y = -\frac{1}{4}x + \frac{1}{4}$, so slope is $-\frac{1}{4}$ (Perpendicular to line $A$).
5. $y = -4x + 6$ ⟶ Slope is $-4$ (Neither parallel nor perpendicular).
6. $y - 4 = 4(x + 3)$ ⟶ This can be rewritten as $y = 4x + 16$, so slope is $4$ (Parallel to line $A$).
7. $y = \frac{1}{4}x + 8$ ⟶ Slope is $\frac{1}{4}$ (Neither parallel nor perpendicular).
8. $y = -\frac{1}{4}x + 9$ ⟶ Slope is $-\frac{1}{4}$ (Perpendicular to line $A$).

Final Sorting:

• Parallel to Line $A$:

  • $y = 4x + 2$
  • $y = 4x + 9$
  • $y - 4 = 4(x + 3)$ ⟶ $y = 4x + 16$

• Perpendicular to Line $A$:

  • $y = -\frac{1}{4}x + 5$
  • $y + 4 = -\frac{1}{4}(x - 3)$ ⟶ $y = -\frac{1}{4}x + \frac{1}{4}$
  • $y = -\frac{1}{4}x + 9$

Would you like any further details or explanations?

Here are some related questions:

1. What are the general conditions for two lines to be perpendicular in 3D space?
2. How does the slope-intercept form of a line differ from the point-slope form?
3. Can lines with different slopes ever be parallel?
4. What are the conditions for lines to be coincident?
5. How does the slope of a line change when reflecting it across the x-axis?

Tip: Always remember, parallel lines share the same slope, and perpendicular lines' slopes multiply to give $-1$.