Two parallel linear equations with slope 3/4, one with y-intercept 1 and the other with y-intercept -3.

The image displays a coordinate plane with two parallel lines, each defined by a linear equation. The task is to solve the problem presented in the image and explain the solution.

Step 1: Analyze the Problem

The image shows two lines:
- Line 1 (purple): $ y = \frac{3}{4}x + 1 $
- Line 2 (blue): $ y = \frac{3}{4}x - 3 $

Both lines have the same slope ($ \frac{3}{4} $), which means they are parallel. The problem likely involves finding the vertical distance between these two parallel lines, as indicated by the dashed arrows connecting points on the lines.

Step 2: Identify Key Points

From the graph:
- Line 1 passes through the point $ (0, 1) $ (y-intercept).
- Line 2 passes through the point $ (0, -3) $ (y-intercept).

The vertical distance between these two points is the difference in their y-coordinates:
$$
\text{Vertical Distance} = |1 - (-3)| = |1 + 3| = 4
$$

Step 3: Verify Using Another Point

To ensure accuracy, let’s verify using another pair of points. For example, when $ x = 4 $:
- For Line 1: $ y = \frac{3}{4}(4) + 1 = 3 + 1 = 4 $. So, the point is $ (4, 4) $.
- For Line 2: $ y = \frac{3}{4}(4) - 3 = 3 - 3 = 0 $. So, the point is $ (4, 0) $.

The vertical distance between these points is:
$$
\text{Vertical Distance} = |4 - 0| = 4
$$

Step 4: General Formula for Distance Between Parallel Lines

For two parallel lines in the form $ y = mx + c_1 $ and $ y = mx + c_2 $, the vertical distance between them is given by:
$$
\text{Distance} = |c_1 - c_2|
$$

Here, $ c_1 = 1 $ and $ c_2 = -3 $, so:
$$
\text{Distance} = |1 - (-3)| = 4
$$

Final Answer

The vertical distance between the two parallel lines $ y = \frac{3}{4}x + 1 $ and $ y = \frac{3}{4}x - 3 $ is 4 units. This is consistent across all corresponding points on the lines, as verified by both graphical observation and algebraic calculation.