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You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Teacher's Edition Go to the latest version.

11.5: Distance and Midpoint Formulas

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

At the end of this lesson, students will be able to:

  • Find the distance between two points in the coordinate plane.
  • Find the missing coordinate of a point given the distance from another known point.
  • Find the midpoint of a line segment.
  • Solve real-world problems using distance and midpoint formulas.


Terms introduced in this lesson:


Teaching Strategies and Tips

Use Examples 1 and 2 to show how the Pythagorean Theorem is used to derive the distance formula.

  • Teachers are encouraged to use a picture in the derivation.

Use Examples 3-5 and Review Questions 7, 8, and 15-20 as thinking problems.

  • Draw pictures to help.
  • Contrast these problems with the mechanical exercises of Example 2 and Review Questions 1-6.

Point out that because of the squares in the distance formula, the order in which the x-values (and the order of the y-values) are plugged in does not matter.

Point out that the midpoint of a segment is found by taking the average values of the x- and y-values of the endpoints.

In Example 9, suggest that students express their answers in radical form.

Error Troubleshooting

General Tip: For points with negative coordinates, remind students about the minus sign in the distance formula.


Find the distance between the two points.

(-2, 5) and (3, -8).

Hint: Plug the values of the two points into the distance formula; notice that parentheses were used around -8.

d = \sqrt{(-2-3)^2 + (5-(-8))^2}

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Date Created:

Feb 22, 2012

Last Modified:

Feb 08, 2016
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