Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.