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To solve this question, we need to determine the characteristics of lines parallel and perpendicular to the line [tex]\( y = \frac{1}{2}x - 4 \)[/tex], passing through the point [tex]\((-4, 2)\)[/tex].
### Slope of the Parallel Line
1. The slope of the given line is [tex]\(\frac{1}{2}\)[/tex].
2. The slope of a line parallel to the given line remains the same. Therefore, the slope of the parallel line is:
[tex]\[ e = \frac{1}{2} \][/tex]
### Equation of the Parallel Line
1. To find the equation of the parallel line passing through [tex]\((-4, 2)\)[/tex], we'll use the slope-intercept form [tex]\( y = mx + c \)[/tex].
2. We know the slope [tex]\( m = \frac{1}{2} \)[/tex] and the point [tex]\((-4, 2)\)[/tex].
3. Substitute the point into the equation to find the y-intercept [tex]\( c \)[/tex]:
[tex]\[ 2 = \frac{1}{2}(-4) + c \][/tex]
[tex]\[ 2 = -2 + c \][/tex]
[tex]\[ c = 4 \][/tex]
Therefore, the equation of the parallel line is:
[tex]\[ y = \frac{1}{2}x + 4 \][/tex]
### A Point on the Parallel Line
1. To find a specific point on this line, let's choose [tex]\( x = 3 \)[/tex]. Substitute [tex]\( x \)[/tex] into the equation:
[tex]\[ y = \frac{1}{2}(3) + 4 \][/tex]
[tex]\[ y = 1.5 + 4 = 5.5 \][/tex]
Thus, a point on the line parallel to the given line and passing through [tex]\((-4, 2)\)[/tex], is:
[tex]\[ (3, 5.5) \][/tex]
### Slope of the Perpendicular Line
1. The slope of the perpendicular line is the negative reciprocal of the slope of the given line.
2. The negative reciprocal of [tex]\( \frac{1}{2} \)[/tex] is:
[tex]\[ -2 \][/tex]
### Equation of the Perpendicular Line
1. Using the slope-intercept form [tex]\( y = mx + c \)[/tex] with the slope [tex]\( m = -2 \)[/tex] and the point [tex]\((-4, 2)\)[/tex]:
2. Substitute the point into the equation to find the y-intercept [tex]\( c \)[/tex]:
[tex]\[ 2 = -2(-4) + c \][/tex]
[tex]\[ 2 = 8 + c \][/tex]
[tex]\[ c = 2 - 8 = -6 \][/tex]
Therefore, the equation of the perpendicular line is:
[tex]\[ y = -2x - 6 \][/tex]
### A Point on the Perpendicular Line
1. To find a specific point on this line, let's choose [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x \)[/tex] into the equation:
[tex]\[ y = -2(0) - 6 \][/tex]
[tex]\[ y = -6 \][/tex]
Thus, a point on the line perpendicular to the given line and passing through [tex]\((-4, 2)\)[/tex], is:
[tex]\[ (0, -6) \][/tex]
### Final Results
- The slope of a line parallel to the given line: [tex]\( e = 0.5 \)[/tex]
- A point on the line parallel to the given line, passing through [tex]\((-4, 2)\)[/tex]: [tex]\( (3, 5.5) \)[/tex]
- The slope of a line perpendicular to the given line: [tex]\( -2.0 \)[/tex]
- A point on the line perpendicular to the given line, passing through [tex]\((-4, 2)\)[/tex]: [tex]\( (0, -6) \)[/tex]
### Slope of the Parallel Line
1. The slope of the given line is [tex]\(\frac{1}{2}\)[/tex].
2. The slope of a line parallel to the given line remains the same. Therefore, the slope of the parallel line is:
[tex]\[ e = \frac{1}{2} \][/tex]
### Equation of the Parallel Line
1. To find the equation of the parallel line passing through [tex]\((-4, 2)\)[/tex], we'll use the slope-intercept form [tex]\( y = mx + c \)[/tex].
2. We know the slope [tex]\( m = \frac{1}{2} \)[/tex] and the point [tex]\((-4, 2)\)[/tex].
3. Substitute the point into the equation to find the y-intercept [tex]\( c \)[/tex]:
[tex]\[ 2 = \frac{1}{2}(-4) + c \][/tex]
[tex]\[ 2 = -2 + c \][/tex]
[tex]\[ c = 4 \][/tex]
Therefore, the equation of the parallel line is:
[tex]\[ y = \frac{1}{2}x + 4 \][/tex]
### A Point on the Parallel Line
1. To find a specific point on this line, let's choose [tex]\( x = 3 \)[/tex]. Substitute [tex]\( x \)[/tex] into the equation:
[tex]\[ y = \frac{1}{2}(3) + 4 \][/tex]
[tex]\[ y = 1.5 + 4 = 5.5 \][/tex]
Thus, a point on the line parallel to the given line and passing through [tex]\((-4, 2)\)[/tex], is:
[tex]\[ (3, 5.5) \][/tex]
### Slope of the Perpendicular Line
1. The slope of the perpendicular line is the negative reciprocal of the slope of the given line.
2. The negative reciprocal of [tex]\( \frac{1}{2} \)[/tex] is:
[tex]\[ -2 \][/tex]
### Equation of the Perpendicular Line
1. Using the slope-intercept form [tex]\( y = mx + c \)[/tex] with the slope [tex]\( m = -2 \)[/tex] and the point [tex]\((-4, 2)\)[/tex]:
2. Substitute the point into the equation to find the y-intercept [tex]\( c \)[/tex]:
[tex]\[ 2 = -2(-4) + c \][/tex]
[tex]\[ 2 = 8 + c \][/tex]
[tex]\[ c = 2 - 8 = -6 \][/tex]
Therefore, the equation of the perpendicular line is:
[tex]\[ y = -2x - 6 \][/tex]
### A Point on the Perpendicular Line
1. To find a specific point on this line, let's choose [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x \)[/tex] into the equation:
[tex]\[ y = -2(0) - 6 \][/tex]
[tex]\[ y = -6 \][/tex]
Thus, a point on the line perpendicular to the given line and passing through [tex]\((-4, 2)\)[/tex], is:
[tex]\[ (0, -6) \][/tex]
### Final Results
- The slope of a line parallel to the given line: [tex]\( e = 0.5 \)[/tex]
- A point on the line parallel to the given line, passing through [tex]\((-4, 2)\)[/tex]: [tex]\( (3, 5.5) \)[/tex]
- The slope of a line perpendicular to the given line: [tex]\( -2.0 \)[/tex]
- A point on the line perpendicular to the given line, passing through [tex]\((-4, 2)\)[/tex]: [tex]\( (0, -6) \)[/tex]
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