Question

determine whether the follwoing lines are parallel, perpendicular, or neither. show your work and explain.

1. $y = \frac{2}{3}x - 4$

$y = \frac{2}{3}x + 1$

1. $y = -5x + 7$

$y = -\frac{1}{3}x - 8$

1. $3x - y = 9$

$x + 3y = 36$

1. $y = 4$

$x = -7$

Explanation:

Problem 97:

Step1: Recall slope-intercept form

The slope - intercept form of a line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.
For the line \(y=\frac{2}{3}x - 4\), the slope \(m_1=\frac{2}{3}\).
For the line \(y=\frac{2}{3}x + 1\), the slope \(m_2=\frac{2}{3}\).

Step2: Determine relationship

Two lines are parallel if their slopes are equal (\(m_1=m_2\)) and they have different y - intercepts (to ensure they are not the same line). Here, \(m_1 = m_2=\frac{2}{3}\) and the y - intercepts are \(- 4\) and \(1\) (different). So the lines are parallel.

Step1: Find slopes

For the line \(y=-5x + 7\), the slope \(m_1=-5\).
For the line \(y =-\frac{1}{3}x-8\), the slope \(m_2=-\frac{1}{3}\).

Step2: Check parallel/ perpendicular

• Parallel: \(m_1

eq m_2\) (\(-5
eq-\frac{1}{3}\)), so not parallel.

• Perpendicular: The product of slopes of two perpendicular lines is \(- 1\). \(m_1\times m_2=(-5)\times(-\frac{1}{3})=\frac{5}{3}

eq - 1\). So neither parallel nor perpendicular.

Step1: Convert to slope - intercept form

• For \(3x - y=9\), solve for \(y\):

\(y = 3x-9\), so slope \(m_1 = 3\).

• For \(x + 3y=36\), solve for \(y\):

\(3y=-x + 36\), \(y=-\frac{1}{3}x + 12\), so slope \(m_2=-\frac{1}{3}\).

Step2: Check perpendicularity

The product of the slopes \(m_1\times m_2=3\times(-\frac{1}{3})=-1\). When the product of two slopes is \(-1\), the lines are perpendicular.

Answer:

Parallel

Problem 98: