Answer :
Certainly! Let's go through the process of finding the equation of the line perpendicular to [tex]\(WX\)[/tex] and passing through the point [tex]\((-1, -2)\)[/tex] step-by-step.
### Step 1: Convert the Equation of [tex]\(WX\)[/tex] to Slope-Intercept Form
The given equation of line [tex]\(WX\)[/tex] is:
[tex]\[ 2x + y = -5 \][/tex]
To convert this to slope-intercept form ([tex]\(y = mx + b\)[/tex]), we need to solve for [tex]\(y\)[/tex]:
[tex]\[ y = -2x - 5 \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of line [tex]\(WX\)[/tex] is [tex]\(-2\)[/tex].
### Step 2: Find the Slope of the Perpendicular Line
The slope of the line perpendicular to another is the negative reciprocal of the original slope. Thus, the slope of the line perpendicular to [tex]\(WX\)[/tex] is:
[tex]\[ \text{Slope}_{\text{perpendicular}} = -\frac{1}{\text{Slope}_{\text{WX}}} = -\frac{1}{-2} = \frac{1}{2} \][/tex]
### Step 3: Use the Point-Slope Form to Find the Equation
We now know the slope of the perpendicular line is [tex]\(\frac{1}{2}\)[/tex], and it passes through the point [tex]\((-1, -2)\)[/tex]. We can use the point-slope form of the equation of a line ([tex]\(y - y_1 = m (x - x_1)\)[/tex]) to find the equation.
Given:
[tex]\[ m = \frac{1}{2}, \quad (x_1, y_1) = (-1, -2) \][/tex]
Substitute these values into the point-slope form:
[tex]\[ y - (-2) = \frac{1}{2}(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 2 = \frac{1}{2} (x + 1) \][/tex]
[tex]\[ y + 2 = \frac{1}{2}x + \frac{1}{2} \][/tex]
### Step 4: Convert to Slope-Intercept Form
To convert this into slope-intercept form ([tex]\(y = mx + b\)[/tex]), we subtract 2 from both sides:
[tex]\[ y = \frac{1}{2}x + \frac{1}{2} - 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
### Step 5: Select the Correct Answer
Thus, the equation of the line perpendicular to [tex]\(WX\)[/tex] and passing through the point [tex]\((-1, -2)\)[/tex] in slope-intercept form is:
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
So, the correct answer is:
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
This corresponds to the third option:
[tex]\[ \boxed{y = \frac{1}{2} x - \frac{3}{2}} \][/tex]
### Step 1: Convert the Equation of [tex]\(WX\)[/tex] to Slope-Intercept Form
The given equation of line [tex]\(WX\)[/tex] is:
[tex]\[ 2x + y = -5 \][/tex]
To convert this to slope-intercept form ([tex]\(y = mx + b\)[/tex]), we need to solve for [tex]\(y\)[/tex]:
[tex]\[ y = -2x - 5 \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of line [tex]\(WX\)[/tex] is [tex]\(-2\)[/tex].
### Step 2: Find the Slope of the Perpendicular Line
The slope of the line perpendicular to another is the negative reciprocal of the original slope. Thus, the slope of the line perpendicular to [tex]\(WX\)[/tex] is:
[tex]\[ \text{Slope}_{\text{perpendicular}} = -\frac{1}{\text{Slope}_{\text{WX}}} = -\frac{1}{-2} = \frac{1}{2} \][/tex]
### Step 3: Use the Point-Slope Form to Find the Equation
We now know the slope of the perpendicular line is [tex]\(\frac{1}{2}\)[/tex], and it passes through the point [tex]\((-1, -2)\)[/tex]. We can use the point-slope form of the equation of a line ([tex]\(y - y_1 = m (x - x_1)\)[/tex]) to find the equation.
Given:
[tex]\[ m = \frac{1}{2}, \quad (x_1, y_1) = (-1, -2) \][/tex]
Substitute these values into the point-slope form:
[tex]\[ y - (-2) = \frac{1}{2}(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 2 = \frac{1}{2} (x + 1) \][/tex]
[tex]\[ y + 2 = \frac{1}{2}x + \frac{1}{2} \][/tex]
### Step 4: Convert to Slope-Intercept Form
To convert this into slope-intercept form ([tex]\(y = mx + b\)[/tex]), we subtract 2 from both sides:
[tex]\[ y = \frac{1}{2}x + \frac{1}{2} - 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
### Step 5: Select the Correct Answer
Thus, the equation of the line perpendicular to [tex]\(WX\)[/tex] and passing through the point [tex]\((-1, -2)\)[/tex] in slope-intercept form is:
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
So, the correct answer is:
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
This corresponds to the third option:
[tex]\[ \boxed{y = \frac{1}{2} x - \frac{3}{2}} \][/tex]