IDNLearn.com makes it easy to find the right answers to your questions. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.

Consider the graph of the line [tex]y = \frac{1}{2}x - 4[/tex] and the point [tex](-4, 2)[/tex].

1. The slope of a line parallel to the given line is [tex]\frac{1}{2}[/tex].

2. A point on the line parallel to the given line, passing through [tex](-4, 2)[/tex], is (your answer here).

3. The slope of a line perpendicular to the given line is [tex]-2[/tex].

4. A point on the line perpendicular to the given line, passing through [tex](-4, 2)[/tex], is (your answer here).


Sagot :

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]