To solve this problem, let’s analyze and address each part:

1. Given Line Equation: The line is given as $y = 5x - 7$.
   • This is in slope-intercept form $y = mx + b$, where $m = 5$. So, the slope of this line is 5.

Part 1: Finding the Equation of the Parallel Line

• For a line to be parallel to another, it must have the same slope. Therefore, the slope of our parallel line will also be 5.
• This parallel line passes through the point $(-7, -3)$.

Using the point-slope form of the equation for a line: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (-7, -3)$ and $m = 5$.

Substituting these values: $y + 3 = 5(x + 7)$ Simplify to get it in slope-intercept form: $y + 3 = 5x + 35$ $y = 5x + 32$ So, the equation of the line parallel to $y = 5x - 7$ and passing through $(-7, -3)$ is: $y = 5x + 32$

Part 2: Finding the Equation of the Perpendicular Line

• For a line to be perpendicular to another, its slope must be the negative reciprocal of the other line’s slope. Since the slope of the original line is 5, the perpendicular slope will be $-\frac{1}{5}$.
• This perpendicular line also passes through the point $(-7, -3)$.

Again, using the point-slope form with $m = -\frac{1}{5}$: $y - (-3) = -\frac{1}{5}(x - (-7))$ Simplify: $y + 3 = -\frac{1}{5}(x + 7)$ Distribute $-\frac{1}{5}$: $y + 3 = -\frac{1}{5}x - \frac{7}{5}$ Subtract 3 from both sides to isolate $y$: $y = -\frac{1}{5}x - \frac{7}{5} - 3$ Convert 3 to a fraction with denominator 5: $y = -\frac{1}{5}x - \frac{7}{5} - \frac{15}{5}$ $y = -\frac{1}{5}x - \frac{22}{5}$ So, the equation of the line perpendicular to $y = 5x - 7$ and passing through $(-7, -3)$ is: $y = -\frac{1}{5}x - \frac{22}{5}$

Final Answers:

• Equation of the parallel line: $y = 5x + 32$
• Equation of the perpendicular line: $y = -\frac{1}{5}x - \frac{22}{5}$

Would you like further explanation on any step or have questions?

Here are some follow-up questions for further exploration:

1. How do you convert a point-slope form equation into slope-intercept form?
2. What is the relationship between the slopes of parallel and perpendicular lines?
3. How can you verify if two lines are perpendicular by looking at their slopes?
4. Why is it important to understand the negative reciprocal when finding perpendicular lines?
5. Can you find another point on each of the lines calculated above?

Tip: Always double-check your calculations by substituting the given point back into the final equation to ensure accuracy.