The problem asks to find the distance between pairs of parallel lines. We need to solve problems 19, 20, 21, 22, 23 and 24.

1. Problem Description

The problem asks to find the distance between pairs of parallel lines. We need to solve problems 19, 20, 21, 22, 23 and

2

4.

2. Solution Steps

1

9. $y = -3$ and $y = 1$. Since these are horizontal lines, the distance is the absolute difference of the $y$-intercepts: $|1 - (-3)| = |1 + 3| = 4$.

2

0. $x = 4$ and $x = -2$. Since these are vertical lines, the distance is the absolute difference of the $x$-intercepts: $|4 - (-2)| = |4 + 2| = 6$.

2

1. $y = 2x + 2$ and $y = 2x - 3$. The distance between two parallel lines $y = mx + b_1$ and $y = mx + b_2$ is given by the formula:

$d = \frac{|b_2 - b_1|}{\sqrt{1 + m^2}}$

In this case, $m = 2$, $b_1 = 2$, and $b_2 = -3$.

$d = \frac{|-3 - 2|}{\sqrt{1 + 2^2}} = \frac{|-5|}{\sqrt{1 + 4}} = \frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}$.

2

2. $y = 4x$ and $y = 4x - 17$. Here, $m = 4$, $b_1 = 0$, and $b_2 = -17$.

$d = \frac{|-17 - 0|}{\sqrt{1 + 4^2}} = \frac{|-17|}{\sqrt{1 + 16}} = \frac{17}{\sqrt{17}} = \frac{17\sqrt{17}}{17} = \sqrt{17}$.

2

3. $y = 2x - 3$ and $2x - y = -4$. Rewrite the second equation as $y = 2x + 4$. Here, $m = 2$, $b_1 = -3$, and $b_2 = 4$.

$d = \frac{|4 - (-3)|}{\sqrt{1 + 2^2}} = \frac{|4 + 3|}{\sqrt{1 + 4}} = \frac{7}{\sqrt{5}} = \frac{7\sqrt{5}}{5}$.

2

4. $y = -\frac{3}{4}x - 1$ and $3x + 4y = 20$. Rewrite the second equation as $4y = -3x + 20$, so $y = -\frac{3}{4}x + 5$. Here, $m = -\frac{3}{4}$, $b_1 = -1$, and $b_2 = 5$.

$d = \frac{|5 - (-1)|}{\sqrt{1 + (-\frac{3}{4})^2}} = \frac{|5 + 1|}{\sqrt{1 + \frac{9}{16}}} = \frac{6}{\sqrt{\frac{16}{16} + \frac{9}{16}}} = \frac{6}{\sqrt{\frac{25}{16}}} = \frac{6}{\frac{5}{4}} = 6 \cdot \frac{4}{5} = \frac{24}{5}$.

3. Final Answer

1

9. 4

2

0. 6

2

1. $\sqrt{5}$

2

2. $\sqrt{17}$

2

3. $\frac{7\sqrt{5}}{5}$

2

4. $\frac{24}{5}$