## Distance Formula Lesson

### What is the Distance Formula?

**The distance formula is a way of finding the distance between two points.** It does this by creating a virtual right triangle and using the Pythagorean theorem. The distance formula has a 2D (two-dimensional) variation and a 3D (three-dimensional) variation.

The 2D distance formula is given as:

$$\begin{align}& d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \end{align}$$

The 3D distance formula is given as:

$$\begin{align}& d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}} \end{align}$$

Where *d* is the distance between the points, (*x _{1}*,

*y*,

_{1}*z*) is point 1, and (

_{1}*x*,

_{2}*y*,

_{2}*z*) is point 2.

_{2}#### How the Distance Formula Works

**The 2D distance formula is the Pythagorean formula applied to two points in the x-y coordinate plane.** The first component inside of the square root is (x_{2} - x_{1})^{2}. This is the horizontal leg of the right triangle. The second component inside the square root is (y_{2} - y_{1})^{2}. This is the vertical leg of the right triangle.

**We can visualize the 3D distance formula as a right triangle that happens to reside in the x-y-z 3D coordinate system.** Because the two points we are measuring between do not sit flat on a 2D plane, we add in the third term with the variable *z*. The third term inside of the square root is (z_{2} - z_{1})^{2}. It allows the distance between the points to be accurately calculated when they are in 3D space.

#### 2D Distance Formula Example Problem

Find the distance between the points (2, 5) and (7, 3).

Solution:

$$\begin{align}& \text{1.) The points lie in a 2D system/plane. So, we will use the 2D formula.} \\ \\ & \text{2.) Let's substitute the points into the equation and then simplify.} \\ \\ & \text{3.) } d = \sqrt{(7 - 2)^{2} + (3 - 5)^{2}} \\ \\ & \text{4.) } d = \sqrt{(5)^{2} + (-2)^{2}} \\ \\ & \text{5.) } d = \sqrt{25 + 4} \\ \\ & \text{6.) The distance between the points is: } \; \boxed{d = \sqrt{29}} \end{align}$$

#### 3D Distance Formula Example Problem

Find the distance between the points (1, 4, 11) and (2, 6, 18).

Solution:

$$\begin{align}& \text{1.) The points are in 3D space, so we will use the 3D distance formula..} \\ \\ & \text{2.) Let's substitute the points into the equation and then simplify.} \\ \\ & \text{3.) } d = \sqrt{(2 - 1)^{2} + (6 - 4)^{2} + (18 - 11)^{2}} \\ \\ & \text{4.) } d = \sqrt{(1)^{2} + (2)^{2} + (7)^{2}} \\ \\ & \text{5.) } d = \sqrt{1 + 4 + 49} \\ \\ & \text{6.) The distance between the points is: } \; \boxed{d = \sqrt{54}} \end{align}$$