Using the Distance Formula
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse.
Figure 14[latex]{c}^{2}={a}^{2}+{b}^{2}\rightarrow c=\sqrt{{a}^{2}+{b}^{2}}[/latex]
It follows that the distance formula is given as
[latex]{d}^{2}={\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}\to d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}[/latex]
We do not have to use the absolute value symbols in this definition because any number squared is positive.
A General Note: The Distance Formula
Given endpoints [latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex], the distance between two points is given by[latex]d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}[/latex]
Example 5: Finding the Distance between Two Points
Find the distance between the points [latex]\left(-3,-1\right)[/latex] and [latex]\left(2,3\right)[/latex].Solution
Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 15.
Figure 15[latex]\begin{array}{l}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\hfill \\ d=\sqrt{{\left(2-\left(-3\right)\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}\hfill \\ =\sqrt{{\left(5\right)}^{2}+{\left(4\right)}^{2}}\hfill \\ =\sqrt{25+16}\hfill \\ =\sqrt{41}\hfill \end{array}[/latex]
Find the distance between two points: [latex]\left(1,4\right)[/latex] and [latex]\left(11,9\right)[/latex].
[latex]\sqrt{125}=5\sqrt{5}[/latex]
Example 6: Finding the Distance between Two Locations
Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.Solution
The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at [latex]\left(1,1\right)[/latex]. The next stop is 5 blocks to the east, so it is at [latex]\left(5,1\right)[/latex]. After that, she traveled 3 blocks east and 2 blocks north to [latex]\left(8,3\right)[/latex]. Lastly, she traveled 4 blocks north to [latex]\left(8,7\right)[/latex]. We can label these points on the grid as in Figure 16.
Figure 16- From her starting location to her first stop at [latex]\left(1,1\right)[/latex], Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.
- Her second stop is at [latex]\left(5,1\right)[/latex]. So from [latex]\left(1,1\right)[/latex] to [latex]\left(5,1\right)[/latex], Tracie drove east 4,000 feet.
- Her third stop is at [latex]\left(8,3\right)[/latex]. There are a number of routes from [latex]\left(5,1\right)[/latex] to [latex]\left(8,3\right)[/latex]. Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.
- Tracie’s final stop is at [latex]\left(8,7\right)[/latex]. This is a straight drive north from [latex]\left(8,3\right)[/latex] for a total of 4,000 feet.
| From/To | Number of Feet Driven |
|---|---|
| [latex]\left(0,0\right)[/latex] to [latex]\left(1,1\right)[/latex] | 2,000 |
| [latex]\left(1,1\right)[/latex] to [latex]\left(5,1\right)[/latex] | 4,000 |
| [latex]\left(5,1\right)[/latex] to [latex]\left(8,3\right)[/latex] | 5,000 |
| [latex]\left(8,3\right)[/latex] to [latex]\left(8,7\right)[/latex] | 4,000 |
| Total | 15,000 |
[latex]\begin{array}{l}d=\sqrt{{\left(8 - 0\right)}^{2}+{\left(7 - 0\right)}^{2}}\hfill \\ =\sqrt{64+49}\hfill \\ =\sqrt{113}\hfill \\ =10.63\text{ units}\hfill \end{array}[/latex]
At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point [latex]\left(8,7\right)[/latex]. Perhaps you have heard the saying "as the crow flies," which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.
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- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
