Questions: Suppose we are given the following. Line 1 passes through (-2,2) and (2,-8). Line 2 passes through (0,-1) and (5,6). Line 3 passes through (4,-7) and (2,-2). (a) Find the slope of each line. Slope of Line 1= Slope of Line 2= Slope of Line 3= (b) For each pair of lines, determine whether they are parallel, perpendicular, or neither. Line 1 and Line 2: Parallel Perpendicular Neither Line 1 and Line 3: Parallel Perpendicular Neither Line 2 and Line 3: Parallel Perpendicular Neither

Solution Steps

To solve this problem, we need to follow these steps:

1. Find the slope of each line: The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(\frac{y_2 - y_1}{x_2 - x_1}\).
2. Determine the relationship between the lines:
   • Two lines are parallel if their slopes are equal.
   • Two lines are perpendicular if the product of their slopes is \(-1\).
   • Otherwise, the lines are neither parallel nor perpendicular.

To find the slopes of the lines, we use the formula for the slope \( m \) given two points \((x_1, y_1)\) and \((x_2, y_2)\):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

For Line 1, passing through points \((-2, 2)\) and \((2, -8)\):

\[ m_1 = \frac{-8 - 2}{2 - (-2)} = \frac{-10}{4} = -2.5 \]

For Line 2, passing through points \((0, -1)\) and \((5, 6)\):

\[ m_2 = \frac{6 - (-1)}{5 - 0} = \frac{7}{5} = 1.4 \]

For Line 3, passing through points \((4, -7)\) and \((2, -2)\):

\[ m_3 = \frac{-2 - (-7)}{2 - 4} = \frac{5}{-2} = -2.5 \]

Next, we analyze the relationships between the lines based on their slopes:

1. Line 1 and Line 2:

   • \( m_1 = -2.5 \)
   • \( m_2 = 1.4 \)
   • Since \( m_1 \neq m_2 \) and \( m_1 \cdot m_2 \neq -1 \), they are Neither parallel nor perpendicular.

2. Line 1 and Line 3:

   • \( m_1 = -2.5 \)
   • \( m_3 = -2.5 \)
   • Since \( m_1 = m_3 \), they are Parallel.

3. Line 2 and Line 3:

   • \( m_2 = 1.4 \)
   • \( m_3 = -2.5 \)
   • Since \( m_2 \neq m_3 \) and \( m_2 \cdot m_3 \neq -1 \), they are Neither parallel nor perpendicular.

Final Answer