Slope of a Line: Practice Questions with Solutions

Practice finding and interpreting the slope of a line with step-by-step solutions. You will classify lines as rising, falling, horizontal, or vertical, and determine whether pairs of lines are parallel, perpendicular, or neither. Slope is the rate of change between two quantities — it appears in speed, cost per unit, and grade/steepness — so mastering it is a core skill in algebra and coordinate geometry.

Slope Formula and Quick Reference

For two distinct points \((x_1, y_1)\) and \((x_2, y_2)\), the slope is

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

  • \(m > 0\): line rises left-to-right.
  • \(m < 0\): line falls left-to-right.
  • \(m = 0\): horizontal line, equation \(y = c\).
  • \(x_2 = x_1\): vertical line, slope undefined, equation \(x = c\).
  • Parallel (non-vertical): \(m_1 = m_2\).
  • Perpendicular (non-vertical): \(m_1 \cdot m_2 = -1\).

Slope Practice Problems

Question 1: Finding and Interpreting Slope

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

  1. \((2, 1)\) and \((4, 5)\)
  2. \((-1, 0)\) and \((3, -5)\)
  3. \((2, 1)\) and \((-3, 1)\)
  4. \((-1, 2)\) and \((-1, -5)\)
View solution
  1. \( m = \dfrac{5 - 1}{4 - 2} = \dfrac{4}{2} = 2 \). Since \(m>0\), the line rises.
  2. \( m = \dfrac{-5 - 0}{3 - (-1)} = \dfrac{-5}{4} = -\dfrac{5}{4} \). Since \(m<0\), the line falls.
  3. \( m = \dfrac{1 - 1}{-3 - 2} = \dfrac{0}{-5} = 0 \). Since \(m = 0\), the line is horizontal.
  4. \( m = \dfrac{-5 - 2}{-1 - (-1)} = \dfrac{-7}{0} \) is undefined, so the line is vertical.

Question 2: Finding a Point Given the Slope

Question: A line has slope \(-2\) and passes through \((2, 5)\). Find another point on the line. (Many answers are possible.)

View solution

Let \((x_1, y_1)\) be any other point on the line. Then

\[ \frac{y_1 - 5}{x_1 - 2} = -2. \]

Choose \(x_1 = -1\) (any value works):

\[ \frac{y_1 - 5}{-3} = -2 \quad\Longrightarrow\quad y_1 - 5 = 6 \quad\Longrightarrow\quad y_1 = 11. \]

One valid answer is \((-1, 11)\). Check: \(\frac{11 - 5}{-1 - 2} = \frac{6}{-3} = -2\) ✓

Question 3: Parallel and Perpendicular Lines

Question: Determine whether \(L_1\) 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. \(L_1\!: (2,-1), (5,-7)\)   \(L_2\!: (0,0), (-1,2)\)
  4. \(L_1\!: (1,0), (2,0)\)   \(L_2\!: (5,-5), (-10,-5)\)
  5. \(L_1\!: (-2,5), (-2,7)\)   \(L_2\!: (5,1), (5,13)\)
View solution

Let \(m_1\) and \(m_2\) be the slopes of \(L_1\) and \(L_2\). Parallel means \(m_1 = m_2\); perpendicular means \(m_1 \cdot m_2 = -1\); otherwise, neither.

  1. \(m_1 = \frac{1-2}{3-1} = -\frac12\), \(m_2 = \frac{0-(-1)}{2-0} = \frac12\). Since \(m_1 \ne m_2\) and \(m_1 m_2 = -\frac14 \ne -1\), the lines are neither.
  2. \(m_1 = \frac{1-3}{3-0} = -\frac23\), \(m_2 = \frac{-5-4}{-7-(-1)} = \frac{-9}{-6} = \frac32\). Since \(m_1 m_2 = -1\), the lines are perpendicular.
  3. \(m_1 = \frac{-7-(-1)}{5-2} = -2\), \(m_2 = \frac{2-0}{-1-0} = -2\). Since \(m_1 = m_2\), the lines are parallel.
  4. \(m_1 = \frac{0-0}{2-1} = 0\), \(m_2 = \frac{-5-(-5)}{-10-5} = 0\). Both are horizontal, so the lines are parallel.
  5. Both slopes are undefined (vertical lines \(x = -2\) and \(x = 5\)), so the lines are parallel.

Special case: a vertical and a horizontal line are perpendicular geometrically, but the slope-product rule \(m_1 m_2 = -1\) does not apply because the vertical line's slope is undefined.

Question 4: Slope Properties

Question: Can two lines with negative slopes be perpendicular?

View solution

No. For two non-vertical lines to be perpendicular, the product of their slopes must be \(-1\). If both slopes are negative, their product is positive, and therefore can never equal \(-1\).

Additional Practice Exercises

Exercise 1: Finding and Interpreting Slope

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

  1. \((3, -1)\) and \((3, 5)\)
  2. \((-1, 0)\) and \((3, 7)\)
  3. \((2, 1)\) and \((6, 0)\)
  4. \((-5, 2)\) and \((9, 2)\)
View solution
  1. \( m = \frac{5 - (-1)}{3 - 3} = \frac{6}{0} \) → undefined → vertical line.
  2. \( m = \frac{7 - 0}{3 - (-1)} = \frac{7}{4} > 0 \) → rising line.
  3. \( m = \frac{0 - 1}{6 - 2} = -\frac{1}{4} < 0 \) → falling line.
  4. \( m = \frac{2 - 2}{9 - (-5)} = 0 \) → horizontal line.

Exercise 2: Find a Point Given the Slope

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

View solution

Let \((x_1, y_1)\) be another point. Then

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

Choose \(y_1 = 11\) (any \(y_1\) works):

\[ \frac{15}{x_1 - 1} = 5 \quad\Longrightarrow\quad x_1 - 1 = 3 \quad\Longrightarrow\quad x_1 = 4. \]

One valid answer is \((4, 11)\). Check: \(\frac{11 - (-4)}{4 - 1} = \frac{15}{3} = 5\) ✓

Exercise 3: Parallel and Perpendicular Lines

Question: Determine whether \(L_1\) 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. \(L_1\!: (1,9), (-2,9)\)   \(L_2\!: (18,-1), (0,-1)\)
View solution
  1. Both slopes undefined (vertical) → parallel.
  2. \(m_1 = -2\), \(m_2 = -\frac12\), \(m_1 m_2 = 1 \ne -1\) → neither.
  3. \(m_1 = -1\), \(m_2 = 1\), \(m_1 m_2 = -1\) → perpendicular.
  4. Both slopes 0 (horizontal) → parallel.

Exercise 4: Slope Properties

Question: Can two lines with positive slopes be perpendicular?

View solution

No. If both slopes are positive, their product is positive and cannot equal \(-1\). Perpendicular non-vertical lines must have slopes of opposite signs.

Additional Resources