**Distance Formula** and Pythagorean Theorem

The distance formula is derived from the Pythagorean theorem. To find the distance between two points (x_{1},y_{1}) and (x_{2},y_{2}), all that you need to do is use the coordinates of these ordered pairs and apply the formula pictured below.

The distance formula is

$
\text{ Distance } = \sqrt{(x_2 -x_1)^2 + (y_2- y_1)^2}
$

Below is a diagram of the distance formula applied to a picture of a line segment

**Video** Tutorial on the Distance Formula

**Practice** Problems

Connect the two points and draw a right triangle. Set up the Pythagorean theorem

$ a^2 + b^2 = \red c^2 \\ 5^2 + 24^2 = \red c^2 $

Now, let's find the distance. .

$ a^2 + b^2 = c^2 \\ 6^2 + 8^2 = c^2 \\ \sqrt{6^2 + 8^2} = c \\ \sqrt{100} = c \\ \fbox{ c = 10} $Note, you could have just plugged the coordinates into the formula, and arrived at the same solution.
The Distance between the points **$$ \boxed {(\blue 6, \red 8) } $$ ** and **$$ \boxed { (\blue 0, \red 0) }$$ **
$
\\ \text{Distance } = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2}
\\ \sqrt{(\blue 0 - \blue 6 )^2 + (\red 0 -\red{ 8 } )^2}
\\ \sqrt{(-6 )^2 + (- 8 )^2}
\\ \sqrt{36 + 64} = \sqrt{100}
\\ \fbox{10}
$

You might be wondering does it matter which $$ \blue x $$ value is $$ \blue{ x_1} $$. For instance, up above we chose $$ \blue {6} $$, from the $$ \boxed {(\blue 6, \red 8) } $$ as $$ \blue {x_1}$$

What if we chose $$ \blue 0 $$ from $$ \boxed { (\blue 0, \red 0) }$$ as $$ \blue {x_1}$$?$ \\ \text{Distance } = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \sqrt{(\blue 6 - \blue 0 )^2 + (\red 8 -\red{ 0 } )^2} \\ \sqrt{(6 )^2 + ( 8 )^2} \\ \sqrt{36 + 64} = \sqrt{100} \\ \fbox{10} $

As you can see it does not matter which $$\blue x$$ value you use first . This is because after you take difference of the $$ \blue x $$ values , you then ** square ** them . And $$ (\red -8)^2 $$ has the same value as $$ (8)^2 $$