# 12.1 A mathematical look at distance

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This module covers mathematical distance in preparation for later modules on conic sections.

The key mathematical formula for discussing all the shapes above is the distance between two points.

Many students are taught, at some point, the “distance formula” as a magic (and very strange-looking) rule. In fact, the distance formula comes directly from a bit of intuition...and the Pythagorean Theorem.

The intuition comes in finding the distance between two points that have one coordinate in common.

## The distance between two points that have one coordinate in common

The drawing shows the points (2,3) and (6,3). Finding the distance between these points is easy: just count! Take your pen and move it along the paper, starting at (2,3) and moving to the right. Let’s see…one unit gets you over to (3,3); the next unit gets you to (4,3)...a couple more units to (6,3). The distance from (2,3) to (6,3) is 4.

Of course, it would be tedious to count our way from (2,3) to (100,3). But we don’t have to—in fact, you may have already guessed the faster way—we subtract the x coordinates.

• The distance from (2,3) to (6,3) is $6-2=4$
• The distance from (2,3) to (100,3) is $100-2=98$

And so on. We can write this generalization in words:

Whenever two points lie on a horizontal line, you can find the distance between them by subtracting their $x$ -coordinates.

This may seem pretty obvious in the examples given above. It’s a little less obvious, but still true, if one of the $x$ coordinates is negative.

The drawing above shows the numbers (-3,1) and (2,1). You can see that the distance between them is 5 (again, by counting). Does our generalization still work? Yes it does, because subtracting a negative number is the same as adding a positive one.

The distance from (-3,1) to (2,1) is $2-\left(-3\right)=5$

How can we express this generalization mathematically? If two points lie on a horizontal line, they have two different x-coordinates: call them ${x}_{1}$ and ${x}_{2}$ . But they have the same y-coordinate, so just call that y. So we can rewrite our generalization like this: “the distance between the points ( ${x}_{1}$ , $y$ ) and ( ${x}_{2}$ , $y$ ) is ${x}_{2}–{x}_{1}$ .” In our most recent example, ${x}_{1}=–3$ , ${x}_{2}=2$ , and $y=1$ . So the generalization says “the distance between the points (-3,1) and (2,1) is $2-\left(-3\right)$ ”, or 5.

But there’s one problem left: what if we had chosen ${x}_{2}$ and ${x}_{1}$ the other way? Then the generalization would say “the distance between the points (2,1) and (-3,1) is $\left(–3\right)-2$ ”, or -5. That isn’t quite right: distances can never be negative. We get around this problem by taking the absolute value of the answer. This guarantees that, no matter what order the points are listed in, the distance will come out positive. So now we are ready for the correct mathematical generalization:

## Distance between two points on a horizontal line

The distance between the points ( ${x}_{1}$ , $y$ ) and ( ${x}_{2}$ , $y$ ) is $|{x}_{2}–{x}_{1}|$

You may want to check this generalization with a few specific examples—try both negative and positive values of ${x}_{1}$ and ${x}_{2}$ . Then, to really test your understanding, write and test a similar generalization for two points that lie on a vertical line together. Both of these results will be needed for the more general case below.

## The distance between two points that have no coordinate in common

So, what if two points have both coordinates different? As an example, consider the distance from (–2,5) to (1,3).

The drawing shows these two points. The (diagonal) line between them has been labeled $d$ : it is this line that we want the length of, since this line represents the distance between our two points.

The drawing also introduces a third point into the picture, the point (–2,3). The three points define the vertices of a right triangle. Based on our earlier discussion, you can see that the vertical line in this triangle is length $|5–3|=2$ . The horizontal line is length $|1–\left(–2\right)|=3$ .

But it is the diagonal line that we want. And we can find that by using the Pythagorean Theorem, which tells us that ${d}^{2}={2}^{2}+{3}^{2}$ . So $d=\sqrt{13}$

If you repeat this process with the generic points ( ${x}_{1}$ , ${y}_{1}$ ) and ( ${x}_{2}$ , ${y}_{2}$ ) you arrive at the distance formula:

## Distance between any two points

If $d$ is the distance between the points ( ${x}_{1}$ , ${y}_{1}$ ) and ( ${x}_{2}$ , ${y}_{1}$ ), then ${d}^{2}=\left({x}_{2}-{x}_{1}{\right)}^{2}+\left({y}_{2}-{y}_{1}{\right)}^{2}$

${x}_{2}–{x}_{1}$ is the horizontal distance, based on our earlier calculation. ${y}_{2}–{y}_{1}$ is the vertical distance, and the entire formula is simply the Pythagorean Theorem restated in terms of coordinates.

And what about those absolute values we had to put in before? They were used to avoid negative distances. Since the distances in the above formulae are being squared , we no longer need the absolute values to insure that all answers will come out positive.

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