Answer :
To determine which lines are perpendicular to the line [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex], we need to find the slope of the given line and then identify the slopes of the other lines given in the question.
1. Find the slope of the given line:
The given line is:
[tex]\[ y - 1 = \frac{1}{3}(x + 2) \][/tex]
We can rewrite this equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 1 = \frac{1}{3}x + \frac{1}{3}(2) \][/tex]
[tex]\[ y - 1 = \frac{1}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} + 1 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{5}{3} \][/tex]
The slope of the given line is [tex]\( \frac{1}{3} \)[/tex].
2. Determine the slope of a line perpendicular to the given line:
For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. Thus, the slope of the perpendicular line should be the negative reciprocal of [tex]\( \frac{1}{3} \)[/tex], which is [tex]\( -3 \)[/tex].
3. Check each given line to see if its slope is [tex]\( -3 \)[/tex]:
- For the line [tex]\( y + 2 = -3(x - 4) \)[/tex]:
[tex]\[ y + 2 = -3(x - 4) \][/tex]
Simplifying, we get:
[tex]\[ y + 2 = -3x + 12 \][/tex]
[tex]\[ y = -3x + 10 \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. Since this matches the slope required for perpendicularity, this line is perpendicular.
- For the line [tex]\( y - 5 = 3(x + 11) \)[/tex]:
[tex]\[ y - 5 = 3(x + 11) \][/tex]
Simplifying, we get:
[tex]\[ y - 5 = 3x + 33 \][/tex]
[tex]\[ y = 3x + 38 \][/tex]
The slope of this line is [tex]\( 3 \)[/tex]. Therefore, this line is not perpendicular.
- For the line [tex]\( y = -3x - \frac{5}{3} \)[/tex]:
[tex]\[ y = -3x - \frac{5}{3} \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. This matches the required slope for perpendicularity, so this line is perpendicular.
- For the line [tex]\( y = \frac{1}{3}x - 2 \)[/tex]:
[tex]\[ y = \frac{1}{3}x - 2 \][/tex]
The slope of this line is [tex]\( \frac{1}{3} \)[/tex]. This does not match the required slope for perpendicularity, so this line is not perpendicular.
- For the line [tex]\( 3x + y = 7 \)[/tex]:
[tex]\[ 3x + y = 7 \][/tex]
Rearranging to slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = -3x + 7 \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. This matches the required slope for perpendicularity, so this line is perpendicular.
Thus, the lines that are perpendicular to [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex] are:
1. [tex]\( y + 2 = -3(x - 4) \)[/tex]
2. [tex]\( y = -3x - \frac{5}{3} \)[/tex]
3. [tex]\( 3x + y = 7 \)[/tex]
The correct lines are [tex]\( 1, 3,\)[/tex] and [tex]\( 5 \)[/tex].
1. Find the slope of the given line:
The given line is:
[tex]\[ y - 1 = \frac{1}{3}(x + 2) \][/tex]
We can rewrite this equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 1 = \frac{1}{3}x + \frac{1}{3}(2) \][/tex]
[tex]\[ y - 1 = \frac{1}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} + 1 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{5}{3} \][/tex]
The slope of the given line is [tex]\( \frac{1}{3} \)[/tex].
2. Determine the slope of a line perpendicular to the given line:
For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. Thus, the slope of the perpendicular line should be the negative reciprocal of [tex]\( \frac{1}{3} \)[/tex], which is [tex]\( -3 \)[/tex].
3. Check each given line to see if its slope is [tex]\( -3 \)[/tex]:
- For the line [tex]\( y + 2 = -3(x - 4) \)[/tex]:
[tex]\[ y + 2 = -3(x - 4) \][/tex]
Simplifying, we get:
[tex]\[ y + 2 = -3x + 12 \][/tex]
[tex]\[ y = -3x + 10 \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. Since this matches the slope required for perpendicularity, this line is perpendicular.
- For the line [tex]\( y - 5 = 3(x + 11) \)[/tex]:
[tex]\[ y - 5 = 3(x + 11) \][/tex]
Simplifying, we get:
[tex]\[ y - 5 = 3x + 33 \][/tex]
[tex]\[ y = 3x + 38 \][/tex]
The slope of this line is [tex]\( 3 \)[/tex]. Therefore, this line is not perpendicular.
- For the line [tex]\( y = -3x - \frac{5}{3} \)[/tex]:
[tex]\[ y = -3x - \frac{5}{3} \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. This matches the required slope for perpendicularity, so this line is perpendicular.
- For the line [tex]\( y = \frac{1}{3}x - 2 \)[/tex]:
[tex]\[ y = \frac{1}{3}x - 2 \][/tex]
The slope of this line is [tex]\( \frac{1}{3} \)[/tex]. This does not match the required slope for perpendicularity, so this line is not perpendicular.
- For the line [tex]\( 3x + y = 7 \)[/tex]:
[tex]\[ 3x + y = 7 \][/tex]
Rearranging to slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = -3x + 7 \][/tex]
The slope of this line is [tex]\( -3 \)[/tex]. This matches the required slope for perpendicularity, so this line is perpendicular.
Thus, the lines that are perpendicular to [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex] are:
1. [tex]\( y + 2 = -3(x - 4) \)[/tex]
2. [tex]\( y = -3x - \frac{5}{3} \)[/tex]
3. [tex]\( 3x + y = 7 \)[/tex]
The correct lines are [tex]\( 1, 3,\)[/tex] and [tex]\( 5 \)[/tex].
