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Determine whether each pair of lines is perpendicular, parallel, or neither.

[tex]\[
\begin{array}{l|l|}
\hline
1. \quad 2y = 4x + 4 & y = -2x - 2 \\
2. \quad 4y = 2x - 4 & y = -2x + 9 \\
3. \quad y = 2x + 4 & \\
\hline
\end{array}
\][/tex]

Neither [tex]$\square$[/tex]

Parallel [tex]$\square$[/tex]

Perpendicular [tex]$\square$[/tex]

Sagot :

Let's determine the relationship between each pair of lines.

### 1. First Pair: [tex]\(2y = 4x + 4\)[/tex] and [tex]\(y = -2x - 2\)[/tex]

To identify the relationship between these lines, we need to express both equations in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.

- For the equation [tex]\(2y = 4x + 4\)[/tex]:

[tex]\[ y = \frac{4x + 4}{2} \\ y = 2x + 2 \][/tex]

So, the slope [tex]\(m_1 = 2\)[/tex].

- For the equation [tex]\(y = -2x - 2\)[/tex]:

[tex]\[ y = -2x - 2 \ (\text{already in slope-intercept form}) \][/tex]

So, the slope [tex]\(m_3 = -2\)[/tex].

Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Here:

[tex]\[ 2 \times -2 = -4 \][/tex]

Thus, these lines are Perpendicular.

### 2. Second Pair: [tex]\(y = -2x - 2\)[/tex] and [tex]\(4y = 2x - 4\)[/tex]

Let's convert both lines to the slope-intercept form.

- For the equation [tex]\(4y = 2x - 4\)[/tex]:

[tex]\[ y = \frac{2x - 4}{4} \\ y = \frac{1}{2}x - 1 \][/tex]

So, the slope is [tex]\(m_2 = \frac{1}{2}\)[/tex].

- For [tex]\(y = -2x - 2\)[/tex]:

[tex]\[ y = -2x - 2 \ (\text{already in slope-intercept form}) \][/tex]

So, the slope [tex]\(m_3 = -2\)[/tex].

The slopes [tex]\(-2\)[/tex] and [tex]\(\frac{1}{2}\)[/tex] do not multiply to [tex]\(-1\)[/tex] and they are not equal, hence these lines are Neither parallel nor perpendicular.

### 3. Third Pair: [tex]\(4y = 2x - 4\)[/tex] and [tex]\(y = -2x + 9\)[/tex]

Express both equations in the slope-intercept form.

- For the equation [tex]\(4y = 2x - 4\)[/tex]:

[tex]\[ y = \frac{2x - 4}{4} \\ y = \frac{1}{2}x - 1 \][/tex]

So, the slope is [tex]\(m_2 = \frac{1}{2}\)[/tex].

- For [tex]\(y = -2x + 9\)[/tex]:

[tex]\[ y = -2x + 9 \ (\text{already in slope-intercept form}) \][/tex]

So, the slope [tex]\(m_4 = -2\)[/tex].

The slopes [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-2\)[/tex] do not multiply to [tex]\(-1\)[/tex] and they are not equal, hence these lines are Neither parallel nor perpendicular.

### 4. Fourth Pair: [tex]\(y = -2x + 9\)[/tex] and [tex]\(y = 2x + 4\)[/tex]

Express both equations in the slope-intercept form.

- For [tex]\(y = -2x + 9\)[/tex]:

[tex]\[ y = -2x + 9 \ (\text{already in slope-intercept form}) \][/tex]

So, the slope [tex]\(m_4 = -2\)[/tex].

- For [tex]\(y = 2x + 4\)[/tex]:

[tex]\[ y = 2x + 4 \ (\text{already in slope-intercept form}) \][/tex]

So, the slope [tex]\(m_5 = 2\)[/tex].

The slopes [tex]\(-2\)[/tex] and [tex]\(2\)[/tex] do not multiply to [tex]\(-1\)[/tex] and they are not equal, hence these lines are Neither parallel nor perpendicular.

Now, let's place the correct relationships in the correct boxes:

- Neither: [tex]\(\square\)[/tex]
- [tex]\(y = -2x - 2\)[/tex] and [tex]\(4y = 2x - 4\)[/tex]
- [tex]\(4y = 2x - 4\)[/tex] and [tex]\(y = -2x + 9\)[/tex]
- [tex]\(y = -2x + 9\)[/tex] and [tex]\(y = 2x + 4\)[/tex]

- Perpendicular: [tex]\(\square\)[/tex]
- [tex]\(2y = 4x + 4\)[/tex] and [tex]\(y = -2x - 2\)[/tex]

So, we have:

Neither:
- [tex]\(y = -2x - 2\)[/tex] and [tex]\(4y = 2x - 4\)[/tex]
- [tex]\(4y = 2x - 4\)[/tex] and [tex]\(y = -2x + 9\)[/tex]
- [tex]\(y = -2x + 9\)[/tex] and [tex]\(y = 2x + 4\)[/tex]

Perpendicular:
- [tex]\(2y = 4x + 4\)[/tex] and [tex]\(y = -2x - 2\)[/tex]