IDNLearn.com offers a comprehensive solution for all your question and answer needs. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
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]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.