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QUESTION IMAGE

select all the equations that represent a line that is perpendicular to…

Question

select all the equations that represent a line that is perpendicular to the line with equation y = 2x - 8.
show your work here
☑️ y = \\(\frac{1}{2}\\)x + 1
☑️ x + 2y = 5

  • x - 2y = 9

○ y = -\\(\frac{1}{2}\\)x + 1
☑️ -x + 2y = -3

Explanation:

To determine which lines are perpendicular to \( y = 2x - 8 \), we first recall that the slope of a line in the form \( y = mx + b \) is \( m \). For two lines to be perpendicular, the product of their slopes must be \( -1 \). The slope of \( y = 2x - 8 \) is \( 2 \), so the slope of a perpendicular line should be \( -\frac{1}{2} \) (since \( 2 \times (-\frac{1}{2}) = -1 \)).

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

The slope here is \( \frac{1}{2} \). The product of \( 2 \) and \( \frac{1}{2} \) is \( 1 \), not \( -1 \). So this line is not perpendicular. (Note: There was a mistake in the initial check; this should not be selected.)

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

The slope is \( -\frac{1}{2} \). The product of \( 2 \) and \( -\frac{1}{2} \) is \( -1 \), so this line is perpendicular.

Step 3: Analyze \( x + 2y = 5 \)

Rewrite in slope - intercept form (\( y = mx + b \)):

$$ LATEXBLOCK0 $$

The slope is \( -\frac{1}{2} \). The product of \( 2 \) and \( -\frac{1}{2} \) is \( -1 \), so this line is perpendicular.

Step 4: Analyze \( -x + 2y = -3 \)

Rewrite in slope - intercept form:

$$ LATEXBLOCK1 $$

The slope is \( \frac{1}{2} \). The product of \( 2 \) and \( \frac{1}{2} \) is \( 1 \), not \( -1 \). So this line is not perpendicular. (Note: There was a mistake in the initial check; this should not be selected.)

Step 5: Analyze \( -x - 2y = 9 \)

Rewrite in slope - intercept form:

$$ LATEXBLOCK2 $$

The slope is \( -\frac{1}{2} \). The product of \( 2 \) and \( -\frac{1}{2} \) is \( -1 \), so this line is perpendicular. (Note: This was not initially checked, but it should be.)

Wait, let's re - do the analysis correctly:

  1. For a line to be perpendicular to \( y = 2x-8 \) (slope \( m_1 = 2 \)), the slope of the perpendicular line \( m_2 \) must satisfy \( m_1\times m_2=-1\), so \( m_2=-\frac{1}{2}\).
Equation 1: \( y=\frac{1}{2}x + 1 \)

Slope \( m=\frac{1}{2}\). \( 2\times\frac{1}{2}=1
eq - 1 \). Not perpendicular.

Equation 2: \( y =-\frac{1}{2}x + 1 \)

Slope \( m =-\frac{1}{2}\). \( 2\times(-\frac{1}{2})=-1 \). Perpendicular.

Equation 3: \( x + 2y=5 \)

Solve for \( y \): \( 2y=-x + 5\Rightarrow y=-\frac{1}{2}x+\frac{5}{2} \). Slope \( m =-\frac{1}{2}\). \( 2\times(-\frac{1}{2})=-1 \). Perpendicular.

Equation 4: \( -x + 2y=-3 \)

Solve for \( y \): \( 2y=x - 3\Rightarrow y=\frac{1}{2}x-\frac{3}{2} \). Slope \( m=\frac{1}{2}\). \( 2\times\frac{1}{2}=1
eq - 1 \). Not perpendicular.

Equation 5: \( -x - 2y=9 \)

Solve for \( y \): \( - 2y=x + 9\Rightarrow y=-\frac{1}{2}x-\frac{9}{2} \). Slope \( m =-\frac{1}{2}\). \( 2\times(-\frac{1}{2})=-1 \). Perpendicular.

So the correct equations are \( y =-\frac{1}{2}x + 1 \), \( x + 2y = 5 \), and \( -x - 2y = 9 \).

Answer:

The equations representing lines perpendicular to \( y = 2x-8 \) are:

  • \( y =-\frac{1}{2}x + 1 \)
  • \( x + 2y = 5 \)
  • \( -x - 2y = 9 \)