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Let's solve the problem step-by-step.

### Given Equation of the Line:
The line equation provided is:
\[
6x + 8y = -8
\]

#### Step 1: Convert the given equation to slope-intercept form (y = mx + b):
First, isolate \(y\).

\[
8y = -6x - 8
\]
\[
y = -\frac{6}{8}x - \frac{8}{8}
\]
\[
y = -\frac{3}{4}x - 1
\]

Thus, the slope of the given line (\(m_1\)) is:
\[
m_1 = -\frac{3}{4}
\]

### Part 1: Equation of the Perpendicular Line
The slope of the line perpendicular to this one will be the negative reciprocal of \(-\frac{3}{4}\). The negative reciprocal of \(-\frac{3}{4}\) is:
\[
m_2 = \frac{4}{3}
\]

Now, using the point \((7, 1)\), we can find the equation of the perpendicular line. Use the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substituting \(m = \frac{4}{3}\), \(x_1 = 7\), and \(y_1 = 1\):
\[
y - 1 = \frac{4}{3}(x - 7)
\]
Simplifying:
\[
y - 1 = \frac{4}{3}x - \frac{28}{3}
\]
\[
y = \frac{4}{3}x - \frac{28}{3} + 1
\]
\[
y = \frac{4}{3}x - \frac{28}{3} + \frac{3}{3}
\]
\[
y = \frac{4}{3}x - \frac{25}{3}
\]

Thus, the equation of the perpendicular line is:
\[
y = \frac{4}{3}x - \frac{25}{3}
\]

### Part 2: Equation of the Parallel Line
The slope of the parallel line will be the same as the slope of the original line, i.e., \(m = -\frac{3}{4}\). Again, using the point \((7, 1)\), we apply the point-slope form:
\[
y - 1 = -\frac{3}{4}(x - 7)
\]
Simplifying:
\[
y - 1 = -\frac{3}{4}x + \frac{21}{4}
\]
\[
y = -\frac{3}{4}x + \frac{21}{4} + 1
\]
\[
y = -\frac{3}{4}x + \frac{21}{4} + \frac{4}{4}
\]
\[
y = -\frac{3}{4}x + \frac{25}{4}
\]

Thus, the equation of the parallel line is:
\[
y = -\frac{3}{4}x + \frac{25}{4}
\]

### Final Answers:
- Equation of the **perpendicular** line: \(y = \frac{4}{3}x - \frac{25}{3}\)
- Equation of the **parallel** line: \(y = -\frac{3}{4}x + \frac{25}{4}\)

Would you like any further clarification on these steps?

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Here are 5 related questions you can explore:
1. How do you find the slope of a line from its equation?
2. What is the geometric meaning of parallel and perpendicular lines?
3. How can you derive the point-slope form of a line?
4. What is the significance of the slope being a negative reciprocal for perpendicular lines?
5. How do you convert between point-slope and slope-intercept forms of a line?

**Tip:** When dealing with perpendicular lines, remember that their slopes are always negative reciprocals of each other.

Consider the line 6x + 8y = -8. Find the equation of the line that is perpendicular to this line and passes through the point (7, 1). Find the equation of the line that is parallel to this line and passes through the point (7, 1).

This problem involves finding the equations of a line that is perpendicular to the given line 6x + 8y = -8 and passes through the point (7, 1), as well as finding a parallel line. Learn how to apply point-slope form and slope-intercept form, along with the concept of negative reciprocal slopes for perpendicular lines.

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