Question

1. match the equation to its perpendicular slope

draggable item → corresponding item
( y = \frac{1}{2}x + 2 ) → ( \frac{1}{2} )
( y = -\frac{1}{2} - 2 ) → ( -2 )
( y = -2x + 1 ) → ( \frac{-1}{2} )
( y = 2x - 1 ) → ( 2 )

Explanation:

To find the perpendicular slope of a line with slope \( m \), we use the property that the product of the slopes of two perpendicular lines is \( -1 \). So, if a line has slope \( m \), the slope of a line perpendicular to it, \( m_{\perp} \), is given by \( m_{\perp}=-\frac{1}{m} \) (for \( m
eq0 \)).

Step 1: For \( y = \frac{1}{2}x + 2 \)

The slope of the given line, \( m=\frac{1}{2} \).
Using the formula for perpendicular slope: \( m_{\perp}=-\frac{1}{\frac{1}{2}}=- 2 \).

Step 2: For \( y=-\frac{1}{2}x - 2 \) (assuming there was a typo and it's \( y = -\frac{1}{2}x-2 \) as the original might have missed \( x \))

The slope of the given line, \( m =-\frac{1}{2} \).
Perpendicular slope: \( m_{\perp}=-\frac{1}{-\frac{1}{2}} = 2 \).

Step 3: For \( y=-2x + 1 \)

The slope of the given line, \( m=-2 \).
Perpendicular slope: \( m_{\perp}=-\frac{1}{-2}=\frac{1}{2} \).

Step 4: For \( y = 2x-1 \)

The slope of the given line, \( m = 2 \).
Perpendicular slope: \( m_{\perp}=-\frac{1}{2} \).

Answer:

• \( y=\frac{1}{2}x + 2 \) corresponds to \( - 2 \)
• \( y = -\frac{1}{2}x-2 \) (corrected) corresponds to \( 2 \)
• \( y=-2x + 1 \) corresponds to \( \frac{1}{2} \)
• \( y = 2x-1 \) corresponds to \( -\frac{1}{2} \)

(Note: If the second equation was indeed \( y = -\frac{1}{2}-2 \) (a horizontal line with slope \( 0 \)), the perpendicular slope would be undefined (vertical line), but the given corresponding items suggest it's a typo and should be \( y=-\frac{1}{2}x - 2 \) to match the context of linear equations with slope -intercept form for the problem of finding perpendicular slopes.)