Skip Navigation

11.5: The Distance and Midpoint Formulas

Difficulty Level: At Grade Created by: CK-12
Turn In

You have already learned you can use the Pythagorean Theorem to understand different types of right triangles and find missing lengths. This lesson will expand its use to include finding the distance between two points on a Cartesian plane.

The Distance Formula

Look at the points on the grid below. Find the length of the segment connecting (1, 5) and (5, 2).

The question asks you to identify the length of the segment. Because the segment is not parallel to either axis, it is difficult to measure given the coordinate grid.

However, it is possible to think of this segment as the hypotenuse of a right triangle. Draw a vertical line and a horizontal line. Find the point of intersection. This point represents the third vertex in the right triangle.

You can easily count the lengths of the legs of this triangle on the grid. The vertical leg extends from (1, 2) to (1, 5), so it is \begin{align*}|5-2|=|3|=3 \ units\end{align*} long. The horizontal leg extends from (1, 2) to (5, 2), so it is \begin{align*}|5-1|=|4| = 4 \ units\end{align*} long. Use the Pythagorean Theorem with these values for the lengths of each leg to find the length of the hypotenuse.

\begin{align*}a^2+b^2&=c^2\\ 3^2+4^2&=c^2\\ 9+16&=c^2\\ 25&=c^2\\ \sqrt{25} & = \sqrt{c^2}\\ 5 & =c\end{align*}

The segment connecting (1, 5) and (5, 2) is 5 units long.

Mathematicians have simplified this process and created a formula that uses these steps to find the distance between any two points in the coordinate plane. If you use the distance formula, you don’t have to draw the extra lines.

Distance Formula: Given points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*}, the length of the segment connecting those two points is \begin{align*}d=\sqrt{(y_2-y_1)^2+(x_2-x_1 )^2}\end{align*}

Example 1: Find the distance between (–3, 5) and (4, –2).

Solution: Use the Distance Formula. Let \begin{align*}(x_1,y_1)=(-3,5)\end{align*} and \begin{align*}(x_2,y_2)=(4,-2)\end{align*}.

\begin{align*}d&=\sqrt{(-2-5)^2+(4-(-3))^2} \rightarrow \sqrt{(-7)^2+7^2}\\ d& =\sqrt{98}=7\sqrt{2} \ units\end{align*}

Example: Point \begin{align*}A=(6, -4)\end{align*} and point \begin{align*}B=(2, k)\end{align*}. What is the value of \begin{align*}k\end{align*} such that the distance between the two points is 5?

Solution: Use the Distance Formula.

\begin{align*}d = \sqrt{(y_1-y_2)^2 + (x_1-x_2)^2} \Rightarrow 5 = \sqrt{(4-k)^2 + (6-2)^2}\end{align*}

\begin{align*}\text{Square both sides of the equation.} \qquad 5^2&= \left [ \sqrt{(4-k)^2 + (6-2)^2} \right ]^2\\ \text{Simplify}. \qquad 25 &= (-4-k)^2 + 16\\ \text{Eliminate the parentheses}. \qquad 0 & = k^2+8k+16 -9\\ \text{Simplify}. \qquad 0 & = k^2+8k+7\\ \text{Find} \ k \ \text{using the Quadratic Formula}. \qquad k&= \frac{-8\pm \sqrt{64-28}}{2}= \frac{-8\pm \sqrt{36}}{2} = \frac{-8 \pm 6}{2}\end{align*}

\begin{align*}k=-7\end{align*} or \begin{align*}k=-1\end{align*}. There are two possibilities for the value of \begin{align*}k\end{align*}. Let’s graph the points to get a visual representation of our results.

Example: At 8 a.m. one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and traveling at a steady rate, Amir was two mile east and four miles north of his starting point. How far did Amir walk and what was his walking speed?

Solution: Plot Amir’s route on a coordinate graph. We can place his starting point at the origin \begin{align*}A=(0, 0)\end{align*}. Then, his ending point will be at point \begin{align*}B=(2, 4)\end{align*}. The distance can be found with the Distance Formula.

\begin{align*}d&=\sqrt{(4-0)^2 + (2-0)^2} = \sqrt{(4)^2 + (2)^2} + \sqrt{16+4}=\sqrt{20}\\ d&=4.47 \ miles.\end{align*}

Since Amir walked 4.47 miles in 2 hours, his speed is:

\begin{align*}\text{Speed} = \frac{4.47 \ miles}{2 \ hours}= 2.24 \ mi/h\end{align*}

The Midpoint Formula

Consider the following situation: You live in Des Moines, Iowa and your grandparents live in Houston, Texas. You plan to visit them for the summer and your parents agree to meet your grandparents halfway to exchange you. How do you find this location?

By meeting something “halfway,” you are finding the midpoint of the straight line connecting the two segments. In the above situation, the midpoint would be halfway between Des Moines and Houston.

The midpoint between two coordinates represents the halfway point, or the average. It is an ordered pair \begin{align*}(x_m,y_m)\end{align*}.

\begin{align*}(x_m,y_m)=\frac{(x_1+x_2)}{2}, \frac{(y_1+ y_2)}{2}\end{align*}

Example: Des Moines, Iowa has the coordinates (41.59, 93.62).

Houston, Texas has the coordinates (29.76, 95.36).

Find the coordinates of the midpoint between these two cities.

Solution: Decide which ordered pair will represent \begin{align*}(x_1,y_1)\end{align*} and which will represent \begin{align*}(x_2,y_2)\end{align*}.

\begin{align*}(x_1,y_1)&=(41.59,93.62)\\ (x_2,y_2)&=(29.76,95.36)\end{align*}

Compute the midpoint using the formula \begin{align*}(x_m,y_m)=\frac{(x_1+x_2)}{2}, \frac{(y_1+ y_2)}{2}\end{align*}

\begin{align*}(x_m,y_m)&=\frac{(41.59+29.76)}{2}, \frac{(93.62+ 95.36)}{2}\\ (x_m,y_m)&=(35.675,94.49)\end{align*}

Using Google Maps, you can meet in the Ozark National Forest, halfway between the two cities.

Example 2: A segment with endpoints (9, –2) and \begin{align*}(x_1,y_1)\end{align*} has a midpoint of (2, –6). Find \begin{align*}(x_1, y_1)\end{align*}.

Solution: Use the Midpoint Formula. \begin{align*}\frac{(x_1+x_2)}{2}=x_m\end{align*}

\begin{align*}2=\frac{x_1+9}{2} \rightarrow 4&=x_1+9\\ x_1&=-5\end{align*}

Following the same \begin{align*}\frac{y_1+(-2)}{2}=-6 \rightarrow y_1+(-2)=-12\end{align*}

\begin{align*}y_1&=-10\\ (x_1,y_1)&=(-5,-10)\end{align*}

Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

CK-12 Basic Algebra: Distance Formula (9:39)

CK-12 Basic Algebra: Midpoint Formula (6:41)

CK-12 Basic Algebra: Visual Pythagorean Theorem Proof (8:50)

CK-12 Basic Algebra: Pythagorean Theorem 3 (3:00)

In 1–10, find the distance between the two points.

  1. \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*}
  2. (7, 7) and (–7, 7)
  3. (–3, 6) and (3, –6)
  4. (–3, –1) and (–5, –8)
  5. (3, –4) and (6, 0)
  6. (–1, 0) and (4, 2)
  7. (–3, 2) and (6, 2)
  8. (0.5, –2.5) and (4, –4)
  9. (12, –10) and (0, –6)
  10. (2.3, 4.5) and (–3.4, –5.2)
  11. Find all points having an \begin{align*}x\end{align*}-coordinate of –4 and whose distance from point (4, 2) is 10.
  12. Find all points having a \begin{align*}y\end{align*}-coordinate of 3 and whose distance from point (–2, 5) is 8.

In 13–22, find the midpoint of the line segment joining the two points.

  1. \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*}
  2. (7, 7) and (–7, 7)
  3. (–3, 6) and (3, –6)
  4. (–3, –1) and (–5, –8)
  5. (3, –4) and (6, 1)
  6. (2, –3) and (2, 4)
  7. (4, –5) and (8, 2)
  8. (1.8, –3.4) and (–0.4, 1.4)
  9. (5, –1) and (–4, 0)
  10. (10, 2) and (2, –4)
  11. An endpoint of a line segment is (4, 5) and the midpoint of the line segment is (3, –2). Find the other endpoint.
  12. An endpoint of a line segment is (–10, –2) and the midpoint of the line segment is (0, 4). Find the other endpoint.
  13. Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles, then the direction of the bike trail changes and she rides in the new direction for a while longer. When she stops, Michelle is 2 miles south and 10 miles west from her starting point. Find the total distance that Michelle covered from her starting point.
  14. Shawn lives six blocks west and ten blocks north of the center of town. Kenya lives fourteen blocks east and two blocks north of the center of town.
    1. How far apart are these two girls “as the crow flies”?
    2. Where is the halfway point between their houses?

Mixed Review

  1. Solve \begin{align*}(x-4)^2=121\end{align*}.
  2. What is the GCF of \begin{align*}21ab^4\end{align*} and \begin{align*}15a^7 b^2\end{align*}?
  3. Evaluate \begin{align*}_{10}C_7\end{align*} and explain its meaning.
  4. Factor \begin{align*}6x^2+17x+5\end{align*}.
  5. Find the area of a rectangle with a length of \begin{align*}(16+2m)\end{align*} and a width of \begin{align*}(12+2m)\end{align*}.
  6. Factor \begin{align*}x^2-81\end{align*}.

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More

Image Attributions

Show Hide Details
Files can only be attached to the latest version of section
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original