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Sagot :
Sure, let's determine the relationship between each pair of lines step-by-step.
### Pair A: [tex]\( y = -\frac{1}{2}x - 12 \)[/tex] and [tex]\( y - 3 = 2(x + 2) \)[/tex]
1. First Line: [tex]\( y = -\frac{1}{2}x - 12 \)[/tex]
- This line is already in the slope-intercept form [tex]\( y = mx + b \)[/tex].
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{1}{2} \)[/tex].
2. Second Line: [tex]\( y - 3 = 2(x + 2) \)[/tex]
- Start by simplifying this equation to the slope-intercept form.
- [tex]\( y - 3 = 2x + 4 \)[/tex]
- Add 3 to both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = 2x + 7 \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( 2 \)[/tex].
3. Comparison:
- The slope of the first line is [tex]\( \frac{-1}{2} \)[/tex].
- The slope of the second line is [tex]\( 2 \)[/tex].
- The product of the slopes [tex]\( -\frac{1}{2} \times 2 = -1 \)[/tex].
Since the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
### Pair B: [tex]\( y - 3 = 6(x + 2) \)[/tex] and [tex]\( y + 3 = -\frac{1}{3}(x - 4) \)[/tex]
1. First Line: [tex]\( y - 3 = 6(x + 2) \)[/tex]
- Simplify to the slope-intercept form.
- [tex]\( y - 3 = 6x + 12 \)[/tex]
- Add 3 to both sides: [tex]\( y = 6x + 15 \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( 6 \)[/tex].
2. Second Line: [tex]\( y + 3 = -\frac{1}{3}(x - 4) \)[/tex]
- Simplify to the slope-intercept form.
- Distribute the [tex]\(-\frac{1}{3}\)[/tex]: [tex]\( y + 3 = -\frac{1}{3}x + \frac{4}{3} \)[/tex]
- Subtract 3 from both sides: [tex]\( y = -\frac{1}{3}x - \frac{5}{3} \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{1}{3} \)[/tex].
3. Comparison:
- The slope of the first line is [tex]\( 6 \)[/tex].
- The slope of the second line is [tex]\( -\frac{1}{3} \)[/tex].
- The product of the slopes [tex]\( 6 \times -\frac{1}{3} = -2 \)[/tex].
Since the product of the slopes is [tex]\(-2\)[/tex], which is not equal to [tex]\(-1\)[/tex], the two lines are neither parallel nor perpendicular.
### Pair C: [tex]\( y = 5 \)[/tex] and [tex]\( x = 5 \)[/tex]
1. First Line: [tex]\( y = 5 \)[/tex]
- This is a horizontal line where for any value of [tex]\( x \)[/tex], [tex]\( y \)[/tex] is always [tex]\( 5 \)[/tex].
- The slope of a horizontal line is [tex]\( 0 \)[/tex].
2. Second Line: [tex]\( x = 5 \)[/tex]
- This is a vertical line where for any value of [tex]\( y \)[/tex], [tex]\( x \)[/tex] is always [tex]\( 5 \)[/tex].
- The slope of a vertical line is undefined (or [tex]\(\infty\)[/tex]).
3. Comparison:
- Horizontal lines and vertical lines are always perpendicular to each other.
Since we have a horizontal line and a vertical line, the two lines are neither parallel nor perpendicular.
So, the relationships between the pairs of lines are:
A. Perpendicular
B. Neither
C. Neither
### Pair A: [tex]\( y = -\frac{1}{2}x - 12 \)[/tex] and [tex]\( y - 3 = 2(x + 2) \)[/tex]
1. First Line: [tex]\( y = -\frac{1}{2}x - 12 \)[/tex]
- This line is already in the slope-intercept form [tex]\( y = mx + b \)[/tex].
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{1}{2} \)[/tex].
2. Second Line: [tex]\( y - 3 = 2(x + 2) \)[/tex]
- Start by simplifying this equation to the slope-intercept form.
- [tex]\( y - 3 = 2x + 4 \)[/tex]
- Add 3 to both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = 2x + 7 \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( 2 \)[/tex].
3. Comparison:
- The slope of the first line is [tex]\( \frac{-1}{2} \)[/tex].
- The slope of the second line is [tex]\( 2 \)[/tex].
- The product of the slopes [tex]\( -\frac{1}{2} \times 2 = -1 \)[/tex].
Since the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
### Pair B: [tex]\( y - 3 = 6(x + 2) \)[/tex] and [tex]\( y + 3 = -\frac{1}{3}(x - 4) \)[/tex]
1. First Line: [tex]\( y - 3 = 6(x + 2) \)[/tex]
- Simplify to the slope-intercept form.
- [tex]\( y - 3 = 6x + 12 \)[/tex]
- Add 3 to both sides: [tex]\( y = 6x + 15 \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( 6 \)[/tex].
2. Second Line: [tex]\( y + 3 = -\frac{1}{3}(x - 4) \)[/tex]
- Simplify to the slope-intercept form.
- Distribute the [tex]\(-\frac{1}{3}\)[/tex]: [tex]\( y + 3 = -\frac{1}{3}x + \frac{4}{3} \)[/tex]
- Subtract 3 from both sides: [tex]\( y = -\frac{1}{3}x - \frac{5}{3} \)[/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{1}{3} \)[/tex].
3. Comparison:
- The slope of the first line is [tex]\( 6 \)[/tex].
- The slope of the second line is [tex]\( -\frac{1}{3} \)[/tex].
- The product of the slopes [tex]\( 6 \times -\frac{1}{3} = -2 \)[/tex].
Since the product of the slopes is [tex]\(-2\)[/tex], which is not equal to [tex]\(-1\)[/tex], the two lines are neither parallel nor perpendicular.
### Pair C: [tex]\( y = 5 \)[/tex] and [tex]\( x = 5 \)[/tex]
1. First Line: [tex]\( y = 5 \)[/tex]
- This is a horizontal line where for any value of [tex]\( x \)[/tex], [tex]\( y \)[/tex] is always [tex]\( 5 \)[/tex].
- The slope of a horizontal line is [tex]\( 0 \)[/tex].
2. Second Line: [tex]\( x = 5 \)[/tex]
- This is a vertical line where for any value of [tex]\( y \)[/tex], [tex]\( x \)[/tex] is always [tex]\( 5 \)[/tex].
- The slope of a vertical line is undefined (or [tex]\(\infty\)[/tex]).
3. Comparison:
- Horizontal lines and vertical lines are always perpendicular to each other.
Since we have a horizontal line and a vertical line, the two lines are neither parallel nor perpendicular.
So, the relationships between the pairs of lines are:
A. Perpendicular
B. Neither
C. Neither
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