Answer :
To determine the relationship between given lines, we need to compare their slopes. Here are the steps to find the slopes and determine if they are parallel, perpendicular, or neither:
### Step 1: Find the slope of each line
Line 1: [tex]\( y = 4x + 8 \)[/tex]
- This equation is already in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- The slope ([tex]\( m \)[/tex]) of Line 1 is [tex]\( 4 \)[/tex].
Line 2: [tex]\( 3x + 12y = -12 \)[/tex]
- To find the slope, we need to rewrite this equation in slope-intercept form [tex]\( y = mx + b \)[/tex].
- Solve for [tex]\( y \)[/tex]:
[tex]\[ 3x + 12y = -12 \][/tex]
[tex]\[ 12y = -3x - 12 \][/tex]
[tex]\[ y = -\frac{1}{4}x - 1 \][/tex]
- The slope ([tex]\( m \)[/tex]) of Line 2 is [tex]\( -\frac{1}{4} \)[/tex].
Line 3: [tex]\( y = 4x - 5 \)[/tex]
- This equation is already in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- The slope ([tex]\( m \)[/tex]) of Line 3 is [tex]\( 4 \)[/tex].
### Step 2: Compare the slopes
Line 1 and Line 2:
- Slope of Line 1: [tex]\( 4 \)[/tex]
- Slope of Line 2: [tex]\( -\frac{1}{4} \)[/tex]
- Two lines are perpendicular if the product of their slopes is [tex]\( -1 \)[/tex].
[tex]\[ 4 \times -\frac{1}{4} = -1 \][/tex]
- Thus, Line 1 and Line 2 are perpendicular.
Line 1 and Line 3:
- Slope of Line 1: [tex]\( 4 \)[/tex]
- Slope of Line 3: [tex]\( 4 \)[/tex]
- Two lines are parallel if they have the same slope.
- Since the slopes are equal, Line 1 and Line 3 are parallel.
Line 2 and Line 3:
- Slope of Line 2: [tex]\( -\frac{1}{4} \)[/tex]
- Slope of Line 3: [tex]\( 4 \)[/tex]
- Two lines are perpendicular if the product of their slopes is [tex]\( -1 \)[/tex].
[tex]\[ -\frac{1}{4} \times 4 = -1 \][/tex]
- Thus, Line 2 and Line 3 are perpendicular.
### Summary:
- Line 1 and Line 2: Perpendicular
- Line 1 and Line 3: Parallel
- Line 2 and Line 3: Perpendicular
So, the final relationships between the lines are:
[tex]\[ \begin{array}{l} \text{Line 1 and Line 2: Perpendicular} \\ \text{Line 1 and Line 3: Parallel} \\ \text{Line 2 and Line 3: Perpendicular} \end{array} \][/tex]
### Step 1: Find the slope of each line
Line 1: [tex]\( y = 4x + 8 \)[/tex]
- This equation is already in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- The slope ([tex]\( m \)[/tex]) of Line 1 is [tex]\( 4 \)[/tex].
Line 2: [tex]\( 3x + 12y = -12 \)[/tex]
- To find the slope, we need to rewrite this equation in slope-intercept form [tex]\( y = mx + b \)[/tex].
- Solve for [tex]\( y \)[/tex]:
[tex]\[ 3x + 12y = -12 \][/tex]
[tex]\[ 12y = -3x - 12 \][/tex]
[tex]\[ y = -\frac{1}{4}x - 1 \][/tex]
- The slope ([tex]\( m \)[/tex]) of Line 2 is [tex]\( -\frac{1}{4} \)[/tex].
Line 3: [tex]\( y = 4x - 5 \)[/tex]
- This equation is already in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- The slope ([tex]\( m \)[/tex]) of Line 3 is [tex]\( 4 \)[/tex].
### Step 2: Compare the slopes
Line 1 and Line 2:
- Slope of Line 1: [tex]\( 4 \)[/tex]
- Slope of Line 2: [tex]\( -\frac{1}{4} \)[/tex]
- Two lines are perpendicular if the product of their slopes is [tex]\( -1 \)[/tex].
[tex]\[ 4 \times -\frac{1}{4} = -1 \][/tex]
- Thus, Line 1 and Line 2 are perpendicular.
Line 1 and Line 3:
- Slope of Line 1: [tex]\( 4 \)[/tex]
- Slope of Line 3: [tex]\( 4 \)[/tex]
- Two lines are parallel if they have the same slope.
- Since the slopes are equal, Line 1 and Line 3 are parallel.
Line 2 and Line 3:
- Slope of Line 2: [tex]\( -\frac{1}{4} \)[/tex]
- Slope of Line 3: [tex]\( 4 \)[/tex]
- Two lines are perpendicular if the product of their slopes is [tex]\( -1 \)[/tex].
[tex]\[ -\frac{1}{4} \times 4 = -1 \][/tex]
- Thus, Line 2 and Line 3 are perpendicular.
### Summary:
- Line 1 and Line 2: Perpendicular
- Line 1 and Line 3: Parallel
- Line 2 and Line 3: Perpendicular
So, the final relationships between the lines are:
[tex]\[ \begin{array}{l} \text{Line 1 and Line 2: Perpendicular} \\ \text{Line 1 and Line 3: Parallel} \\ \text{Line 2 and Line 3: Perpendicular} \end{array} \][/tex]