Select the correct answer.

Are the given lines parallel, perpendicular, or neither?

[tex]\[
\begin{array}{l}
3x + 12y = 9 \\
2x - 8y = 4
\end{array}
\][/tex]

A. The quotient of the slopes of the lines is 1, so the lines are parallel.

B. The product of the slopes of the lines is 1, so the lines are perpendicular.

C. The slopes of the lines are not the same, so they are neither.



Answer :

Alright class, let's take a detailed look at the given equations and determine the relationship between the two lines.

We start with the equations:

1. [tex]\(3x + 12y = 9\)[/tex]
2. [tex]\(2x - 8y = 4\)[/tex]

To understand the relationship, we'll need to find the slopes of these lines. We'll convert each equation to slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope.

### Step-by-Step Solution:

#### For the first equation: [tex]\(3x + 12y = 9\)[/tex]

1. Isolate [tex]\(y\)[/tex]:
[tex]\[ 12y = -3x + 9 \][/tex]
2. Divide every term by 12:
[tex]\[ y = -\frac{3}{12}x + \frac{9}{12} \][/tex]
3. Simplify the fractions:
[tex]\[ y = -\frac{1}{4}x + \frac{3}{4} \][/tex]

So, the slope [tex]\(m_1\)[/tex] of the first line is [tex]\(-\frac{1}{4}\)[/tex].

#### For the second equation: [tex]\(2x - 8y = 4\)[/tex]

1. Isolate [tex]\(y\)[/tex]:
[tex]\[ -8y = -2x + 4 \][/tex]
2. Divide every term by [tex]\(-8\)[/tex]:
[tex]\[ y = \frac{-2}{-8}x + \frac{4}{-8} \][/tex]
3. Simplify the fractions:
[tex]\[ y = \frac{1}{4}x - \frac{1}{2} \][/tex]

So, the slope [tex]\(m_2\)[/tex] of the second line is [tex]\(\frac{1}{4}\)[/tex].

### Analyzing the Slopes

Now that we have the slopes:
- [tex]\(m_1 = -\frac{1}{4}\)[/tex]
- [tex]\(m_2 = \frac{1}{4}\)[/tex]

Let's check if the lines are either parallel or perpendicular:

1. Parallel Lines: Two lines are parallel if their slopes are equal. Here, [tex]\(-\frac{1}{4}\)[/tex] is not equal to [tex]\(\frac{1}{4}\)[/tex], so the lines are not parallel.

2. Perpendicular Lines: Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Let's calculate the product of the slopes:
[tex]\[ \left( -\frac{1}{4} \right) \times \left( \frac{1}{4} \right) = -\frac{1}{16} \][/tex]

Since [tex]\(-\frac{1}{16}\)[/tex] is not equal to [tex]\(-1\)[/tex], the lines are also not perpendicular.

### Conclusion

Since the slopes of the lines are not equal, the lines are not parallel. Furthermore, the product of their slopes is [tex]\(-\frac{1}{16}\)[/tex] rather than [tex]\(-1\)[/tex], so they are not perpendicular either.

Hence, the lines are neither parallel nor perpendicular.