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Sagot :
To determine whether the given lines are parallel, perpendicular, or neither, we first need to find the slopes of each line.
The given equations are:
1. [tex]\( 3x + 12y = 9 \)[/tex]
2. [tex]\( 2x - 8y = 4 \)[/tex]
First, we rearrange each equation into the slope-intercept form ([tex]\( y = mx + b \)[/tex]), where [tex]\( m \)[/tex] represents the slope:
### For the first equation:
[tex]\[ 3x + 12y = 9 \][/tex]
1. Isolate [tex]\( y \)[/tex]:
[tex]\[ 12y = -3x + 9 \][/tex]
2. Divide by 12 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{3}{12}x + \frac{9}{12} \][/tex]
[tex]\[ y = -\frac{1}{4}x + \frac{3}{4} \][/tex]
So, the slope ([tex]\( m_1 \)[/tex]) of the first line is:
[tex]\[ m_1 = -\frac{1}{4} \][/tex]
### For the second equation:
[tex]\[ 2x - 8y = 4 \][/tex]
1. Isolate [tex]\( y \)[/tex]:
[tex]\[ -8y = -2x + 4 \][/tex]
2. Divide by -8 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{2}{8}x - \frac{4}{8} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{1}{2} \][/tex]
So, the slope ([tex]\( m_2 \)[/tex]) of the second line is:
[tex]\[ m_2 = \frac{1}{4} \][/tex]
Now that we have the slopes:
- Slope of the first line, [tex]\( m_1 = -\frac{1}{4} \)[/tex]
- Slope of the second line, [tex]\( m_2 = \frac{1}{4} \)[/tex]
Next, let's determine the relationship between the two lines:
1. Parallel: Lines are parallel if their slopes are equal.
- [tex]\( m_1 \neq m_2 \)[/tex] so the lines are not parallel.
2. Perpendicular: Lines are perpendicular if the product of their slopes is -1.
- [tex]\( m_1 \times m_2 = -\frac{1}{4} \times \frac{1}{4} = -\frac{1}{16} \neq -1 \)[/tex] so the lines are not perpendicular.
3. Neither: Since the slopes are not equal and their product is not -1, the lines are neither parallel nor perpendicular.
Therefore, the correct answer is: the lines are neither parallel nor perpendicular.
The given equations are:
1. [tex]\( 3x + 12y = 9 \)[/tex]
2. [tex]\( 2x - 8y = 4 \)[/tex]
First, we rearrange each equation into the slope-intercept form ([tex]\( y = mx + b \)[/tex]), where [tex]\( m \)[/tex] represents the slope:
### For the first equation:
[tex]\[ 3x + 12y = 9 \][/tex]
1. Isolate [tex]\( y \)[/tex]:
[tex]\[ 12y = -3x + 9 \][/tex]
2. Divide by 12 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{3}{12}x + \frac{9}{12} \][/tex]
[tex]\[ y = -\frac{1}{4}x + \frac{3}{4} \][/tex]
So, the slope ([tex]\( m_1 \)[/tex]) of the first line is:
[tex]\[ m_1 = -\frac{1}{4} \][/tex]
### For the second equation:
[tex]\[ 2x - 8y = 4 \][/tex]
1. Isolate [tex]\( y \)[/tex]:
[tex]\[ -8y = -2x + 4 \][/tex]
2. Divide by -8 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{2}{8}x - \frac{4}{8} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{1}{2} \][/tex]
So, the slope ([tex]\( m_2 \)[/tex]) of the second line is:
[tex]\[ m_2 = \frac{1}{4} \][/tex]
Now that we have the slopes:
- Slope of the first line, [tex]\( m_1 = -\frac{1}{4} \)[/tex]
- Slope of the second line, [tex]\( m_2 = \frac{1}{4} \)[/tex]
Next, let's determine the relationship between the two lines:
1. Parallel: Lines are parallel if their slopes are equal.
- [tex]\( m_1 \neq m_2 \)[/tex] so the lines are not parallel.
2. Perpendicular: Lines are perpendicular if the product of their slopes is -1.
- [tex]\( m_1 \times m_2 = -\frac{1}{4} \times \frac{1}{4} = -\frac{1}{16} \neq -1 \)[/tex] so the lines are not perpendicular.
3. Neither: Since the slopes are not equal and their product is not -1, the lines are neither parallel nor perpendicular.
Therefore, the correct answer is: the lines are neither parallel nor perpendicular.
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