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# 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.

## Vocabulary

Terms introduced in this lesson:

distance
equidistant
midpoint

## 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\begin{align*}x-\end{align*}values (and the order of the y\begin{align*}y-\end{align*}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\begin{align*}x-\end{align*} and y\begin{align*}y-\end{align*}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.

Example:

Find the distance between the two points.

(2,5)\begin{align*}(-2, 5)\end{align*} and (3,8)\begin{align*}(3, -8)\end{align*}.

Hint: Plug the values of the two points into the distance formula; notice that parentheses were used around 8\begin{align*}-8\end{align*}.

d=(23)2+(5(8))2\begin{align*}d = \sqrt{(-2-3)^2 + (5-(-8))^2}\end{align*}

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Date Created:
Feb 22, 2012
Last Modified:
Jun 22, 2016
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CK.MAT.ENG.TE.1.Algebra-I.11.5
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