To determine whether the given pairs of lines are parallel, perpendicular, or neither, we need to find the slopes of each line.

1. Line 1: $3y = -5x + 5$

   • Solving for $y$: $y = -\frac{5}{3}x + \frac{5}{3}$
   • The slope of Line 1 is $m_1 = -\frac{5}{3}$.

2. Line 2: $y = -\frac{5}{3}x - 8$

   • The equation is already in slope-intercept form ($y = mx + b$), so the slope of Line 2 is $m_2 = -\frac{5}{3}$.

3. Line 3: $6x + 10y = 8$

   • Solving for $y$: $10y = -6x + 8$ $y = -\frac{3}{5}x + \frac{4}{5}$
   • The slope of Line 3 is $m_3 = -\frac{3}{5}$.

Analyzing the Pairs:

• Line 1 and Line 2: Both lines have the same slope $-\frac{5}{3}$, so they are parallel.
• Line 1 and Line 3: The slopes are $-\frac{5}{3}$ and $-\frac{3}{5}$. The product of these slopes is: $\left(-\frac{5}{3}\right) \times \left(-\frac{3}{5}\right) = 1$ Since the product is 1, these lines are perpendicular.
• Line 2 and Line 3: Similarly, the slopes are $-\frac{5}{3}$ and $-\frac{3}{5}$, making these lines perpendicular as well.

Final Answers:

• Line 1 and Line 2: Parallel
• Line 1 and Line 3: Perpendicular
• Line 2 and Line 3: Perpendicular

Would you like further details on the steps involved or have any questions?

Here are some related questions to expand on this:

1. How can you determine if two lines are parallel by looking at their equations?
2. What is the significance of the product of slopes when finding perpendicular lines?
3. Can lines with different y-intercepts still be parallel?
4. How would you approach this problem if the equations were given in a different form, such as point-slope form?
5. Are there cases where two lines can be neither parallel nor perpendicular but still intersect?

Tip: Remember, parallel lines have identical slopes, while perpendicular lines have slopes that multiply to $-1$.