To determine which lines are perpendicular to the line [tex]\(y - 1 = \frac{1}{3}(x + 2)\)[/tex], we need to first rewrite this line in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept.
Starting with the given equation:
[tex]\[ y - 1 = \frac{1}{3}(x + 2) \][/tex]
Distribute [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ y - 1 = \frac{1}{3}x + \frac{2}{3} \][/tex]
Add 1 to both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} + 1 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} + \frac{3}{3} \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{5}{3} \][/tex]
So, the slope [tex]\(m\)[/tex] of the given line is [tex]\(\frac{1}{3}\)[/tex].
Lines that are perpendicular to this line will have slopes that are the negative reciprocals of [tex]\(\frac{1}{3}\)[/tex]. The negative reciprocal of [tex]\(\frac{1}{3}\)[/tex] is [tex]\(-3\)[/tex].
Now, we will check each of the given lines to see if any have a slope of [tex]\(-3\)[/tex]:
1. Line [tex]\( y + 2 = -3(x - 4) \)[/tex]
[tex]\[ y + 2 = -3(x - 4) \][/tex]
Distribute [tex]\(-3\)[/tex]:
[tex]\[ y + 2 = -3x + 12 \][/tex]
Subtract 2 from both sides:
[tex]\[ y = -3x + 10 \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(-3\)[/tex]. This line is perpendicular to the given line.
2. Line [tex]\( y - 5 = 3(x + 11) \)[/tex]
[tex]\[ y - 5 = 3(x + 11) \][/tex]
Distribute [tex]\(3\)[/tex]:
[tex]\[ y - 5 = 3x + 33 \][/tex]
Add 5 to both sides:
[tex]\[ y = 3x + 38 \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(3\)[/tex]. This line is not perpendicular to the given line.
3. Line [tex]\( y = -3x - \frac{5}{3} \)[/tex]
This line is already in the slope-intercept form:
[tex]\[ y = -3x - \frac{5}{3} \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(-3\)[/tex]. This line is perpendicular to the given line.
4. Line [tex]\( y = \frac{1}{3}x - 2 \)[/tex]
This line is already in the slope-intercept form:
[tex]\[ y = \frac{1}{3}x - 2 \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(\frac{1}{3}\)[/tex]. This line is not perpendicular to the given line.
5. Line [tex]\( 3x + y = 7 \)[/tex]
Rewrite in slope-intercept form:
[tex]\[ y = -3x + 7 \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(-3\)[/tex]. This line is perpendicular to the given line.
In 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]
Therefore, the lines that are perpendicular are the first, third, and fifth lines on the list:
- [tex]\( y + 2 = -3(x - 4) \)[/tex]
- [tex]\( y = -3x - \frac{5}{3} \)[/tex]
- [tex]\( 3x + y = 7 \)[/tex]
