To determine which lines are perpendicular to the given line [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex], we need to follow these steps:
### Step 1: Find the Slope of the Given Line
Let's start by determining the slope of the given line. The equation [tex]\( y - 1 = \frac{1}{3}(x + 2) \)[/tex] can be rewritten in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] denotes the slope.
First, expand and simplify the equation:
[tex]\[ y - 1 = \frac{1}{3}x + \frac{2}{3} \][/tex]
Add 1 to both sides:
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} + 1 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{5}{3} \][/tex]
Thus, the slope of the given line ([tex]\( m \)[/tex]) is [tex]\( \frac{1}{3} \)[/tex].
### Step 2: Perpendicular Line Slope
Perpendicular lines have slopes that are negative reciprocals of one another. Therefore, if the slope of the given line is [tex]\( \frac{1}{3} \)[/tex], the slope of any line perpendicular to it will be the negative reciprocal:
[tex]\[ \text{slope of perpendicular line} = -\frac{1}{ \left( \frac{1}{3} \right)} = -3 \][/tex]
### Step 3: Determine Slopes of the Given Lines
Now we need to find the slopes of the given lines to see which ones have a slope of [tex]\( -3 \)[/tex]:
1. Line: [tex]\( y + 2 = -3(x - 4) \)[/tex]
[tex]\[ y + 2 = -3x + 12 \][/tex]
Subtract 2 from both sides:
[tex]\[ y = -3x + 10 \][/tex]
The slope is [tex]\( -3 \)[/tex].
2. Line: [tex]\( y - 5 = 3(x + 11) \)[/tex]
[tex]\[ y - 5 = 3x + 33 \][/tex]
Add 5 to both sides:
[tex]\[ y = 3x + 38 \][/tex]
The slope is [tex]\( 3 \)[/tex].
3. Line: [tex]\( y = -3x - \frac{5}{3} \)[/tex]
The equation is already in slope-intercept form.
The slope is [tex]\( -3 \)[/tex].
4. Line: [tex]\( y = \frac{1}{3}x - 2 \)[/tex]
The equation is already in slope-intercept form.
The slope is [tex]\( \frac{1}{3} \)[/tex].
5. Line: [tex]\( 3x + y = 7 \)[/tex]
Rewrite in the slope-intercept form:
[tex]\[ y = -3x + 7 \][/tex]
The slope is [tex]\( -3 \)[/tex].
### Step 4: Identify Perpendicular Lines
Based on our calculations, the lines with slopes of [tex]\( -3 \)[/tex] are:
- [tex]\( y + 2 = -3(x - 4) \)[/tex]
- [tex]\( y = -3x - \frac{5}{3} \)[/tex]
- [tex]\( 3x + y = 7 \)[/tex]
Thus, 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]
