Question

(04.02 mc) the equation of line wx is 2x + y = -5. what is the equation of a line perpendicular to line wx in slope - intercept form that contains point (-1, -2)?
options:
( y = \frac{1}{2}x + \frac{3}{2} )
( y = -\frac{1}{2}x + \frac{3}{2} )
( y = \frac{1}{2}x - \frac{3}{2} )
( y = -\frac{1}{2}x - \frac{3}{2} )

Explanation:

Step1: Find slope of line WX

Given line \( 2x + y = -5 \), rewrite in slope - intercept form \( y=mx + b \) (where \( m \) is slope).
Subtract \( 2x \) from both sides: \( y=-2x - 5 \). So slope of WX, \( m_1=-2 \).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \( - 1 \). Let slope of perpendicular line be \( m_2 \).
Then \( m_1\times m_2=-1 \). Substitute \( m_1 = - 2 \): \( -2\times m_2=-1 \).
Solve for \( m_2 \): \( m_2=\frac{-1}{-2}=\frac{1}{2} \).

Step3: Use point - slope form to find equation

Point - slope form is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(-1,-2) \) and \( m = \frac{1}{2} \).
Substitute values: \( y-(-2)=\frac{1}{2}(x - (-1)) \).
Simplify: \( y + 2=\frac{1}{2}(x + 1) \).

Step4: Convert to slope - intercept form

Expand right - hand side: \( y + 2=\frac{1}{2}x+\frac{1}{2} \).
Subtract 2 from both sides: \( y=\frac{1}{2}x+\frac{1}{2}-2 \).
Simplify \( \frac{1}{2}-2=\frac{1 - 4}{2}=-\frac{3}{2} \)? Wait, no, wait. Wait, \( 2=\frac{4}{2} \), so \( \frac{1}{2}-\frac{4}{2}=-\frac{3}{2} \)? Wait, no, wait the point is \( (-1,-2) \). Wait, let's re - calculate step 3 and 4.
Wait, point - slope: \( y - y_1=m(x - x_1) \), \( x_1=-1 \), \( y_1 = - 2 \), \( m=\frac{1}{2} \).
So \( y-(-2)=\frac{1}{2}(x-(-1))\Rightarrow y + 2=\frac{1}{2}(x + 1) \).
Expand: \( y+2=\frac{1}{2}x+\frac{1}{2} \).
Subtract 2 (which is \( \frac{4}{2} \)) from both sides: \( y=\frac{1}{2}x+\frac{1}{2}-\frac{4}{2}=\frac{1}{2}x-\frac{3}{2} \)? Wait, no, the options have \( y=\frac{1}{2}x+\frac{3}{2} \), \( y =-\frac{1}{2}x+\frac{3}{2} \), \( y=\frac{1}{2}x-\frac{3}{2} \), \( y =-\frac{1}{2}x-\frac{3}{2} \). Wait, maybe I made a mistake in sign. Wait, the point is \( (-1,-2) \). Let's re - do step 3.
Wait, \( y - y_1=m(x - x_1) \), \( y_1=-2 \), so \( y-(-2)=y + 2 \), \( x_1=-1 \), so \( x-(-1)=x + 1 \). So \( y + 2=\frac{1}{2}(x + 1) \).
\( y=\frac{1}{2}x+\frac{1}{2}-2=\frac{1}{2}x+\frac{1 - 4}{2}=\frac{1}{2}x-\frac{3}{2} \)? No, wait the options:
Wait the options are:
1. \( y=\frac{1}{2}x+\frac{3}{2} \)
2. \( y =-\frac{1}{2}x+\frac{3}{2} \)
3. \( y=\frac{1}{2}x-\frac{3}{2} \)
4. \( y =-\frac{1}{2}x-\frac{3}{2} \)

Wait, maybe I messed up the sign of the slope. Wait, no, the slope of WX is \( - 2 \), so the perpendicular slope is \( \frac{1}{2} \) (since \( (-2)\times\frac{1}{2}=-1 \)). Now, let's use the point \( (-1,-2) \) again.
\( y=mx + b \), \( m=\frac{1}{2} \), so \( y=\frac{1}{2}x + b \). Substitute \( x=-1 \), \( y = - 2 \):
\( -2=\frac{1}{2}(-1)+b \)
\( -2=-\frac{1}{2}+b \)
Add \( \frac{1}{2} \) to both sides: \( b=-2+\frac{1}{2}=-\frac{4}{2}+\frac{1}{2}=-\frac{3}{2} \)? No, that's not matching the first option. Wait, maybe the point is \( (-1,2) \)? Wait, the user's image shows the point as \( (-1, - 2) \)? Wait, maybe I misread the point. Wait, the problem says "contains point (-1, - 2)"? Wait, no, maybe the point is \( (-1,2) \)? Let's check with \( (-1,2) \).
If \( (x_1,y_1)=(-1,2) \), then \( y - 2=\frac{1}{2}(x + 1) \)
\( y-2=\frac{1}{2}x+\frac{1}{2} \)
\( y=\frac{1}{2}x+\frac{1}{2}+2=\frac{1}{2}x+\frac{1 + 4}{2}=\frac{1}{2}x+\frac{5}{2} \), no. Wait, maybe the original problem's point is \( (-1,1) \)? No, the options have \( y=\frac{1}{2}x+\frac{3}{2} \). Let's check with \( m=\frac{1}{2} \) and the first option \( y=\frac{1}{2}x+\frac{3}{2} \). Plug in \( x=-1 \): \( y=\frac{1}{2}(-1)+\frac{3}{2}=\frac{-1 + 3}{2}=1 \). Not - 2. Second option: \( y=-\frac{1}{2}x+\frac{3}{2} \), \( x = - 1 \), \( y=\frac{1}{2}+\frac{3}{2}=2 \). Third option: \( y=\frac{1}{2}x-\frac{3}{2} \), \( x=-1 \), \( y=-\frac{1}{2}-\frac{3}{2}=-2 \). Ah! So the point is \( (-1,-2) \), and the equation is \( y=\frac{1}{2}x-\frac{3}{2} \), which is the third option (the option with \( y=\frac{1}{2}x-\frac{3}{2} \), looking at the original options: the third option in the image (from left to right) is \( y=\frac{1}{2}x-\frac{3}{2} \)? Wait, the original options:
First option: \( y=\frac{1}{2}x+\frac{3}{2} \)
Second option: \( y =-\frac{1}{2}x+\frac{3}{2} \)
Third option: \( y=\frac{1}{2}x-\frac{3}{2} \)
Fourth option: \( y =-\frac{1}{2}x-\frac{3}{2} \)

So when we plug \( x = - 1 \) into \( y=\frac{1}{2}x-\frac{3}{2} \), we get \( y=\frac{1}{2}(-1)-\frac{3}{2}=-\frac{1}{2}-\frac{3}{2}=-2 \), which matches the point \( (-1,-2) \).

Answer:

The equation of the perpendicular line is \( y=\frac{1}{2}x-\frac{3}{2} \) (the option corresponding to \( y=\frac{1}{2}x-\frac{3}{2} \), which is the third option in the given choices).