The problem asks us to find the distance between each pair of parallel lines for problems 19-24.

1. Problem Description

The problem asks us to find the distance between each pair of parallel lines for problems 19-

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4.

2. Solution Steps

Problem 19:

The equations are $y = -3$ and $y = 1$. These are horizontal lines. The distance between them is the absolute difference in their y-intercepts.

Distance $= |1 - (-3)| = |1 + 3| = 4$

Problem 20:

The equations are $x = 4$ and $x = -2$. These are vertical lines. The distance between them is the absolute difference in their x-intercepts.

Distance $= |4 - (-2)| = |4 + 2| = 6$

Problem 21:

The equations are $y = 2x + 2$ and $y = 2x - 3$. These lines have the same slope ($m = 2$), so they are parallel. To find the distance between them, we can use the formula:

$d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$

where the equations are in the form $ax + by + c = 0$. Rewriting the equations:

$2x - y + 2 = 0$ and $2x - y - 3 = 0$.

Then $a = 2$, $b = -1$, $c_1 = 2$, and $c_2 = -3$.

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

Problem 22:

The equations are $y = 4x$ and $y = 4x - 17$. These lines have the same slope ($m = 4$), so they are parallel. Rewriting the equations:

$4x - y = 0$ and $4x - y - 17 = 0$.

Then $a = 4$, $b = -1$, $c_1 = 0$, and $c_2 = -17$.

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

Problem 23:

The equations are $y = 2x - 3$ and $2x - y = -4$, which we can rewrite as $y = 2x + 4$. These lines have the same slope ($m=2$), so they are parallel.

Rewriting the equations: $2x - y - 3 = 0$ and $2x - y + 4 = 0$.

Then $a = 2$, $b = -1$, $c_1 = -3$, and $c_2 = 4$.

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

Problem 24:

The equations are $y = -\frac{3}{4}x - 1$ and $3x + 4y = 20$. Rewrite the second equation as $4y = -3x + 20$, so $y = -\frac{3}{4}x + 5$. These lines have the same slope ($m = -\frac{3}{4}$), so they are parallel.

Rewrite the equations: $3x + 4y + 4 = 0$ and $3x + 4y - 20 = 0$.

Then $a = 3$, $b = 4$, $c_1 = 4$, and $c_2 = -20$.

$d = \frac{|-20 - 4|}{\sqrt{3^2 + 4^2}} = \frac{|-24|}{\sqrt{9 + 16}} = \frac{24}{\sqrt{25}} = \frac{24}{5}$

3. Final Answer

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9. 4

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0. 6

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1. $\sqrt{5}$

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2. $\sqrt{17}$

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

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