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To determine which lines are perpendicular to the line [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex], we need to first understand the concept of the slope and the property of perpendicular lines.
### Step 1: Determine the slope of the given line
The equation of the line is given in point-slope form:
[tex]\[ y - 1 = \frac{1}{3}(x + 2) \][/tex]
The slope-intercept form of a line is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. From the equation [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex], we can see that the slope [tex]\( m \)[/tex] is [tex]\( \frac{1}{3} \)[/tex].
### Step 2: Find the perpendicular slope
Lines that are perpendicular have slopes that are negative reciprocals of each other. The negative reciprocal of [tex]\( \frac{1}{3} \)[/tex] is [tex]\( -3 \)[/tex].
### Step 3: Identifying the slopes of the given lines
Now let's find the slopes of the given lines and check which of them has a slope of [tex]\( -3 \)[/tex]:
1. Line 1: [tex]\( y + 2 = -3(x - 4) \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y + 2 = -3x + 12 \)[/tex]
- Subtract 2 from both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = -3x + 10 \)[/tex]
- Slope: [tex]\( -3 \)[/tex]
2. Line 2: [tex]\( y - 5 = 3(x + 11) \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y - 5 = 3x + 33 \)[/tex]
- Add 5 to both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = 3x + 38 \)[/tex]
- Slope: [tex]\( 3 \)[/tex]
3. Line 3: [tex]\( y = -3x - \frac{5}{3} \)[/tex]
- This line is already in slope-intercept form.
- Slope: [tex]\( -3 \)[/tex]
4. Line 4: [tex]\( y = \frac{1}{3}x - 2 \)[/tex]
- This line is already in slope-intercept form.
- Slope: [tex]\( \frac{1}{3} \)[/tex]
5. Line 5: [tex]\( 3x + y = 7 \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y = -3x + 7 \)[/tex]
- Slope: [tex]\( -3 \)[/tex]
### Step 4: Compare slopes
We are looking for lines with a slope of [tex]\( -3 \)[/tex], which is the negative reciprocal of [tex]\( \frac{1}{3} \)[/tex].
- Line 1: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
- Line 2: Slope = [tex]\( 3 \)[/tex] (Not Perpendicular)
- Line 3: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
- Line 4: Slope = [tex]\( \frac{1}{3} \)[/tex] (Not Perpendicular)
- Line 5: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
### Conclusion
The lines that are perpendicular to [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex] are:
[tex]\[ y + 2 = -3(x - 4) \][/tex]
[tex]\[ y = -3x - \frac{5}{3} \][/tex]
[tex]\[ 3x + y = 7 \][/tex]
These correspond to Line 1, Line 3, and Line 5 respectively. Hence, the lines that are perpendicular to the given line are:
[tex]\[ [1, 3, 5] \][/tex]
### Step 1: Determine the slope of the given line
The equation of the line is given in point-slope form:
[tex]\[ y - 1 = \frac{1}{3}(x + 2) \][/tex]
The slope-intercept form of a line is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. From the equation [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex], we can see that the slope [tex]\( m \)[/tex] is [tex]\( \frac{1}{3} \)[/tex].
### Step 2: Find the perpendicular slope
Lines that are perpendicular have slopes that are negative reciprocals of each other. The negative reciprocal of [tex]\( \frac{1}{3} \)[/tex] is [tex]\( -3 \)[/tex].
### Step 3: Identifying the slopes of the given lines
Now let's find the slopes of the given lines and check which of them has a slope of [tex]\( -3 \)[/tex]:
1. Line 1: [tex]\( y + 2 = -3(x - 4) \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y + 2 = -3x + 12 \)[/tex]
- Subtract 2 from both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = -3x + 10 \)[/tex]
- Slope: [tex]\( -3 \)[/tex]
2. Line 2: [tex]\( y - 5 = 3(x + 11) \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y - 5 = 3x + 33 \)[/tex]
- Add 5 to both sides to solve for [tex]\( y \)[/tex]: [tex]\( y = 3x + 38 \)[/tex]
- Slope: [tex]\( 3 \)[/tex]
3. Line 3: [tex]\( y = -3x - \frac{5}{3} \)[/tex]
- This line is already in slope-intercept form.
- Slope: [tex]\( -3 \)[/tex]
4. Line 4: [tex]\( y = \frac{1}{3}x - 2 \)[/tex]
- This line is already in slope-intercept form.
- Slope: [tex]\( \frac{1}{3} \)[/tex]
5. Line 5: [tex]\( 3x + y = 7 \)[/tex]
- Rewrite in slope-intercept form: [tex]\( y = -3x + 7 \)[/tex]
- Slope: [tex]\( -3 \)[/tex]
### Step 4: Compare slopes
We are looking for lines with a slope of [tex]\( -3 \)[/tex], which is the negative reciprocal of [tex]\( \frac{1}{3} \)[/tex].
- Line 1: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
- Line 2: Slope = [tex]\( 3 \)[/tex] (Not Perpendicular)
- Line 3: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
- Line 4: Slope = [tex]\( \frac{1}{3} \)[/tex] (Not Perpendicular)
- Line 5: Slope = [tex]\( -3 \)[/tex] (Perpendicular)
### Conclusion
The lines that are perpendicular to [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex] are:
[tex]\[ y + 2 = -3(x - 4) \][/tex]
[tex]\[ y = -3x - \frac{5}{3} \][/tex]
[tex]\[ 3x + y = 7 \][/tex]
These correspond to Line 1, Line 3, and Line 5 respectively. Hence, the lines that are perpendicular to the given line are:
[tex]\[ [1, 3, 5] \][/tex]
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