Answer :
Sure! Let's go through a detailed, step-by-step solution:
We have two lines given by the equations:
[tex]\[ \begin{array}{l} 3x + 12y = 9 \\ 2x - 8y = 4 \end{array} \][/tex]
First, let's convert these equations into slope-intercept form, [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope of the line.
1. For the first line [tex]\(3x + 12y = 9\)[/tex]:
[tex]\[ \begin{aligned} 3x + 12y &= 9 \\ 12y &= -3x + 9 \\ y &= \frac{-3}{12}x + \frac{9}{12} \\ y &= -\frac{1}{4}x + \frac{3}{4} \end{aligned} \][/tex]
The slope [tex]\(m_1\)[/tex] of the first line is [tex]\(-\frac{1}{4}\)[/tex].
2. For the second line [tex]\(2x - 8y = 4\)[/tex]:
[tex]\[ \begin{aligned} 2x - 8y &= 4 \\ -8y &= -2x + 4 \\ y &= \frac{2}{8}x - \frac{4}{8} \\ y &= \frac{1}{4}x - \frac{1}{2} \end{aligned} \][/tex]
The slope [tex]\(m_2\)[/tex] of the second line is [tex]\(\frac{1}{4}\)[/tex].
Now, let's analyze the relationship between the two slopes [tex]\(m_1 = -\frac{1}{4}\)[/tex] and [tex]\(m_2 = \frac{1}{4}\)[/tex]:
- The two slopes are [tex]\(-\frac{1}{4}\)[/tex] and [tex]\(\frac{1}{4}\)[/tex].
Checking whether the lines are parallel, perpendicular, or neither involves:
1. Parallel Lines: Two lines are parallel if their slopes are equal, i.e., [tex]\(m_1 = m_2\)[/tex].
[tex]\[ -\frac{1}{4} \ne \frac{1}{4} \][/tex]
Hence, the lines are not parallel.
2. Perpendicular Lines: Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex], i.e., [tex]\(m_1 \cdot m_2 = -1\)[/tex].
[tex]\[ -\frac{1}{4} \times \frac{1}{4} = -\frac{1}{16} \ne -1 \][/tex]
Hence, the lines are not perpendicular.
3. Neither Parallel nor Perpendicular: If the slopes are not equal and their product is not [tex]\(-1\)[/tex], then the lines are neither parallel nor perpendicular.
Since [tex]\(-\frac{1}{4} \neq \frac{1}{4}\)[/tex] and their product [tex]\(-\frac{1}{16} \ne -1\)[/tex], the lines are neither parallel nor perpendicular.
The correct answer is:
- The slopes of the lines are opposites, so they are neither parallel nor perpendicular.
Thus, the solution to the problem is:
[tex]\[ \boxed{3} \][/tex]
We have two lines given by the equations:
[tex]\[ \begin{array}{l} 3x + 12y = 9 \\ 2x - 8y = 4 \end{array} \][/tex]
First, let's convert these equations into slope-intercept form, [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope of the line.
1. For the first line [tex]\(3x + 12y = 9\)[/tex]:
[tex]\[ \begin{aligned} 3x + 12y &= 9 \\ 12y &= -3x + 9 \\ y &= \frac{-3}{12}x + \frac{9}{12} \\ y &= -\frac{1}{4}x + \frac{3}{4} \end{aligned} \][/tex]
The slope [tex]\(m_1\)[/tex] of the first line is [tex]\(-\frac{1}{4}\)[/tex].
2. For the second line [tex]\(2x - 8y = 4\)[/tex]:
[tex]\[ \begin{aligned} 2x - 8y &= 4 \\ -8y &= -2x + 4 \\ y &= \frac{2}{8}x - \frac{4}{8} \\ y &= \frac{1}{4}x - \frac{1}{2} \end{aligned} \][/tex]
The slope [tex]\(m_2\)[/tex] of the second line is [tex]\(\frac{1}{4}\)[/tex].
Now, let's analyze the relationship between the two slopes [tex]\(m_1 = -\frac{1}{4}\)[/tex] and [tex]\(m_2 = \frac{1}{4}\)[/tex]:
- The two slopes are [tex]\(-\frac{1}{4}\)[/tex] and [tex]\(\frac{1}{4}\)[/tex].
Checking whether the lines are parallel, perpendicular, or neither involves:
1. Parallel Lines: Two lines are parallel if their slopes are equal, i.e., [tex]\(m_1 = m_2\)[/tex].
[tex]\[ -\frac{1}{4} \ne \frac{1}{4} \][/tex]
Hence, the lines are not parallel.
2. Perpendicular Lines: Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex], i.e., [tex]\(m_1 \cdot m_2 = -1\)[/tex].
[tex]\[ -\frac{1}{4} \times \frac{1}{4} = -\frac{1}{16} \ne -1 \][/tex]
Hence, the lines are not perpendicular.
3. Neither Parallel nor Perpendicular: If the slopes are not equal and their product is not [tex]\(-1\)[/tex], then the lines are neither parallel nor perpendicular.
Since [tex]\(-\frac{1}{4} \neq \frac{1}{4}\)[/tex] and their product [tex]\(-\frac{1}{16} \ne -1\)[/tex], the lines are neither parallel nor perpendicular.
The correct answer is:
- The slopes of the lines are opposites, so they are neither parallel nor perpendicular.
Thus, the solution to the problem is:
[tex]\[ \boxed{3} \][/tex]
