**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 $$ (8)^2 $$

**Graph the two points**

Plug in the side lengths into the **Pythagoren Theorem**

Note, it does not matter which 'way' you draw the right triangle. It can be either up above or down below.

$ a^2 + b^2 = c^2 \\ 5^2 + 24^2 = \red c^2 $**Solve** for the Hyptenuse

Note, you could have just plugged the coordinates into the formula, and arrived at the same solution.

Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula.

The Distance between the points $$(\blue 2, \red 4) \text{ and } ( \blue{ 26} , \red 9)$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} = \sqrt{(\blue 2 -\blue { 26} )^2 + (\red 4 - \red{9} )^2} \\ \text{d} = \sqrt{(\blue {-24 })^2 + (\red{- 5} )^2} \\ \color{green}{ \text{d} = \sqrt{576 + 25 }} \\ \boxed{ \text{d} =\text{distance} = \sqrt{601}= 24.5 } $**Graph the two points**

Plug in the side lengths into the **Pythagoren Theorem**

**Solve** for the Hyptenuse

Note, you could have just plugged the coordinates into the formula, and arrived at the same solution.

Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula.

The Distance between the points $$(\blue 4, \red 6) \text{ and } ( \blue{ 28} , \red {13} )$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} =\sqrt{(\blue 4 -\blue { 28} )^2 + (\red 6 - \red{ 13 } )^2} \\ \text{d} = \sqrt{(\blue {-24 })^2 + (\red{-7} )^2} \\ \text{d} = \color{green}{ \sqrt{576 + 49 }} \\ \boxed{ \text{d} = \sqrt{625}= 25 } $**Graph the two points**

Plug in the side lengths into the **Pythagoren Theorem**

**Solve** for the Hyptenuse

Note, you could have just plugged the coordinates into the formula, and arrived at the same solution.

Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula.

The Distance between the points $$(\blue 4, \red 8 ) \text{ and } ( \blue{ 12} , \red {14} )$$ $ \\ \text{d} = \sqrt{(\blue {x_2} -\blue{x_1})^2 + (\red{ y_2} - \red{ y_1})^2} \\ \text{d} = \sqrt{(\blue{12} -\blue { 4} )^2 + (\red 14 - \red{8} )^2} \\ \text{d} = \sqrt{(\blue {8} )^2 + (\red{6} )^2} \\ \color{green}{ \text{d} = \sqrt{64 + 36 }} \\ \text{d} = \sqrt{100 } \\ \boxed{ \text{d} =\text{d} =10 } $