The Distance Formula

The Distance Formula 

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The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Here's how we get from the one to the other:

Suppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The points look like this:
You can draw in the lines that form a right-angled triangle, using these points as two of the corners:
It's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values:

Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle):

c² = a² + b²

...so:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

This format always holds true. Given two points, you can always plot them, draw the right triangle, and then find the length of the hypotenuse. The length of the hypotenuse is the distance between the two points. Since this format always works, it can be turned into a formula:

Distance Formula: Given the two points (x₁, y₁) and (x₂, y₂), the distance between these points is given by the formula:

Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The distance will be the same, regardless.

• Find the distance between the points (–2, –3) and (–4, 4).

I just plug the coordinates into the Distance Formula:

Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.
The most common mistake made when using the Formula is to accidentally mismatch the x-values and y-values. Be careful you don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly.
Also, don't get careless with the square-root symbol. If you get in the habit of omitting the square root and then "remembering" to put it back in when you check your answers in the back of the book, then you'll forget the square root on the test, and you'll miss easy points.
You also don't want to be careless with the squaring inside the Formula. Remember that you simplify inside the parentheses before you square, not after, and remember that the square is on everything inside the parentheses, including the minus sign, so the square of a negative is a positive.
By the way, it is almost always better to leave the answer in "exact" form (the square root "" above). Rounding is usually reserved for the last step of word problems. If you're not sure which format is preferred, do both, like this:

Now YOU try it!

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Very often you will encounter the Distance Formula in veiled forms. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. For instance:

• Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.

The radius is the distance between the center and any point on the circle, so I need to find the distance: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

Then the radius is sqrt(10), or about 3.16, rounded to two decimal places.

• Find all points (4, y) that are 10 units from the point (–2, –1).
I'll plug the two points and the distance into the Distance Formula:

Now I'll square both sides, so I can get to the variable:

This means y = –9 or y = 7, so the two points are (4, –9) and (4, 7).

If you're not sure why there are two points that solve this exercise, try drawing the (–2, –1) and then drawing a circle with radius 10 around this. Then draw the vertical line through x = 4. You'll see that the vertical line crosses the circle in two spots: (4, –9) and (4, 7).