# Two Ways to Find Distance on the XY-Coordinate Plane: Which Is Better for You?

When the SAT asks us to find the distance between two points on the* xy*-coordinate plane, there are **always two ways to do it**: with the Distance Formula and with the Pythagorean Theorem, which are actually two variations of the same formula. Most students find one of the methods easier, and it helps to **walk in on test day knowing which method works faster for you**. Here’s the sample question we’ll be working with in this post:

What’s the distance between the two points?

#### Method 1: Distance Formula

The nice thing about the Distance Formula is that you just plug in the coordinates of the two points you want to find the distance between. The less-nice thing is that it’s quite a formula:

[math]d=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}[/math]

Let’s use it on the problem we’ve got:

[math]d=\sqrt{(-5-5)^{2}+(5-{-3})^{2}}[/math] [math]d=\sqrt{(-10)^{2}+(8)^{2}}[/math] [math]d=\sqrt{100+64}[/math] [math]d=\sqrt{164}[/math] [math]d\approx{12.8}[/math]

#### Method 2: Pythagorean Theorem

The nice thing about this method is that you already have the formula memorized. The downside is that you have to take a few steps before you use it.

**Step 1: Draw a right triangle**

Using the points as the two ends of the hypotenuse, draw a right triangle. It doesn’t matter which way we draw it: at right, for example, the method would still work if we drew the triangle above the hypotenuse instead of beneath it.

**Step 2: Calculate leg lengths & take absolute value**

To find the leg lengths, we just** find the difference** between the 2 *x*-coordinates and then between the 2 *y*-coordinates. Since distances can’t be negative, we always **take the absolute value** of the difference to plug into the Theorem. In this case, we would calculate:

*x-*coordinates: [math]-5-5=-10[/math]
horizontal leg = 10 units

*y-*coordinates: [math]5-(-3)=8[/math]
vertical leg = 8 units

**Step 3: Plug lengths into Pythagorean Theorem**

Then we plug the leg lengths into the formula, and voila!

[math]a^{2}+b^{2}=c^{2}[/math] [math]8^{2}+10^{2}=c^{2}[/math] [math]64+100=c^{2}[/math] [math]\sqrt{164}=c[/math] [math]12.8\approx{c}[/math]

So now that you know these problems can always be solved two ways, which do you think will be easier and faster for you? Once you decide, practice that method until it becomes automatic. Then you’ll be ready to speed through distance questions and spend your time on the *really* hard ones.