Slope of Lines: Practice Questions with Step-by-Step Solutions

Master the concept of slope with these practice problems and detailed solutions. Learn to identify rising, falling, horizontal, and vertical lines, and understand relationships between parallel and perpendicular lines.

Slope Practice Problems with Solutions

Question 1: Finding and Interpreting Slope

Find the slope of the line passing through each pair of points and describe the line as rising, falling, horizontal, or vertical.

  1. \( (2, 1), (4, 5) \)
  2. \( (-1, 0), (3, -5) \)
  3. \( (2, 1), (-3, 1) \)
  4. \( (-1, 2), (-1, -5) \)

Solution to Question 1

  1. The slope is calculated as:

    \( m = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2 \)

    Since \( m > 0 \), the line rises as \( x \) increases.

  2. The slope is calculated as:

    \( m = \frac{-5 - 0}{3 - (-1)} = \frac{-5}{4} = -\frac{5}{4} \)

    Since \( m < 0 \), the line falls as \( x \) increases.

  3. The slope is calculated as:

    \( m = \frac{1 - 1}{-3 - 2} = \frac{0}{-5} = 0 \)

    Since \( m = 0 \), the line is horizontal (parallel to the x-axis).

  4. The slope is calculated as:

    \( m = \frac{-5 - 2}{-1 - (-1)} = \frac{-7}{0} \)

    Division by zero is undefined, so the slope is undefined and the line is vertical (parallel to the y-axis).

Question 2: Finding a Point Given Slope

A line has a slope of \( -2 \) and passes through the point \( (2, 5) \). Find another point that lies on this line. (Multiple solutions possible)

Solution to Question 2

Let \( (x_1, y_1) \) be the coordinates of the unknown point. Using the slope formula:

\( \frac{y_1 - 5}{x_1 - 2} = -2 \)

This equation has infinite solutions. Choosing \( x_1 = -1 \):

\( \frac{y_1 - 5}{-1 - 2} = -2 \)

\( \frac{y_1 - 5}{-3} = -2 \)

\( y_1 - 5 = 6 \)

\( y_1 = 11 \)

One possible point is \( (-1, 11) \).

Verification: \( \frac{11 - 5}{-1 - 2} = \frac{6}{-3} = -2 \) ✓

Question 3: Parallel and Perpendicular Lines

Determine whether lines \( L_1 \) and \( L_2 \) are parallel, perpendicular, or neither.

  1. \( L_1: (1, 2), (3, 1) \)
    \( L_2: (0, -1), (2, 0) \)
  2. \( L_1: (0, 3), (3, 1) \)
    \( L_2: (-1, 4), (-7, -5) \)
  3. \( L_2: (2, -1), (5, -7) \)
    \( L_2: (0, 0), (-1, 2) \)
  4. \( L_1: (1, 0), (2, 0) \)
    \( L_2: (5, -5), (-10, -5) \)
  5. \( L_1: (-2, 5), (-2, 7) \)
    \( L_2: (5, 1), (5, 13) \)

Solution to Question 3

Let \( m_1 \) = slope of \( L_1 \), \( m_2 \) = slope of \( L_2 \).

  1. \( m_1 = \frac{1 - 2}{3 - 1} = -\frac{1}{2} \), \( m_2 = \frac{0 - (-1)}{2 - 0} = \frac{1}{2} \)

    \( m_1 \neq m_2 \) and \( m_1 \times m_2 \neq -1 \), so lines are neither parallel nor perpendicular.

  2. \( m_1 = \frac{1 - 3}{3 - 0} = -\frac{2}{3} \), \( m_2 = \frac{-5 - 4}{-7 - (-1)} = \frac{-9}{-6} = \frac{3}{2} \)

    \( m_1 \times m_2 = -\frac{2}{3} \times \frac{3}{2} = -1 \), so lines are perpendicular.

  3. \( m_1 = \frac{-7 - (-1)}{5 - 2} = \frac{-6}{3} = -2 \), \( m_2 = \frac{2 - 0}{-1 - 0} = -2 \)

    \( m_1 = m_2 \), so lines are parallel.

  4. \( m_1 = \frac{0 - 0}{2 - 1} = 0 \), \( m_2 = \frac{-5 - (-5)}{-10 - 5} = \frac{0}{-15} = 0 \)

    \( m_1 = m_2 = 0 \), so both lines are horizontal and parallel.

  5. \( m_1 = \frac{7 - 5}{-2 - (-2)} = \frac{2}{0} \) (undefined), \( m_2 = \frac{13 - 1}{5 - 5} = \frac{12}{0} \) (undefined)

    Both slopes are undefined, so both lines are vertical and parallel.

Question 4: Slope Properties

Can two lines with negative slopes be perpendicular?

Solution to Question 4

No. For perpendicular lines, \( m_1 \times m_2 = -1 \). If both slopes are negative, their product would be positive, never -1.

Additional Practice Exercises

Find the slope of the line through each pair of points and describe the line.

  1. \( (3, -1), (3, 5) \)
  2. \( (-1, 0), (3, 7) \)
  3. \( (2, 1), (6, 0) \)
  4. \( (-5, 2), (9, 2) \)

Exercise 2: Find a Point Given Slope

A line has slope \( 5 \) and passes through \( (1, -4) \). Find another point on this line.

Exercise 3: Parallel and Perpendicular Lines

Determine if lines \( L_1 \) and \( L_2 \) are parallel, perpendicular, or neither.

  1. \( L_1: (1, 2), (1, 1) \)
    \( L_2: (-4, -1), (-4, 0) \)
  2. \( L_1: (2, 3), (3, 1) \)
    \( L_2: (1, -2), (7, -5) \)
  3. \( L_1: (1, -1), (2, -2) \)
    \( L_2: (0, 0), (1, 1) \)
  4. \( L_1: (1, 9), (-2, 9) \)
    \( L_2: (18, -1), (0, -1) \)

Exercise 4: Slope Properties

Can two lines with positive slopes be perpendicular?

Exercise Solutions

Solution to Exercise 1

  1. \( m = \frac{5 - (-1)}{3 - 3} = \frac{6}{0} \) (undefined, vertical line)
  2. \( m = \frac{7 - 0}{3 - (-1)} = \frac{7}{4} \) (positive, rising line)
  3. \( m = \frac{0 - 1}{6 - 2} = -\frac{1}{4} \) (negative, falling line)
  4. \( m = \frac{2 - 2}{9 - (-5)} = 0 \) (zero, horizontal line)

Solution to Exercise 2

Using slope formula with point \( (x_1, y_1) \):

\( \frac{y_1 - (-4)}{x_1 - 1} = 5 \)

Choosing \( y_1 = 11 \):

\( \frac{11 + 4}{x_1 - 1} = 5 \)

\( \frac{15}{x_1 - 1} = 5 \)

\( 15 = 5(x_1 - 1) \)

\( x_1 = 4 \)

One possible point: \( (4, 11) \).

Solution to Exercise 3

  1. \( m_1 = \frac{1 - 2}{1 - 1} = \frac{-1}{0} \) (undefined), \( m_2 = \frac{0 - (-1)}{-4 - (-4)} = \frac{1}{0} \) (undefined). Both vertical, so parallel.
  2. \( m_1 = \frac{1 - 3}{3 - 2} = -2 \), \( m_2 = \frac{-5 - (-2)}{7 - 1} = -\frac{1}{2} \). Product \( = 1 \neq -1 \), so neither.
  3. \( m_1 = \frac{-2 - (-1)}{2 - 1} = -1 \), \( m_2 = \frac{1 - 0}{1 - 0} = 1 \). Product \( = -1 \), so perpendicular.
  4. \( m_1 = \frac{9 - 9}{-2 - 1} = 0 \), \( m_2 = \frac{-1 - (-1)}{0 - 18} = 0 \). Both horizontal, so parallel.

Solution to Exercise 4

No. For perpendicular lines, \( m_1 \times m_2 = -1 \). If both slopes are positive, their product is positive, never -1.

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