The Distance Formula

Distance Formula and Pythagorean Theorem

The distance formula is derived from the Pythagorean theorem. To find the distance between two points (x1,y1) and (x2,y2), 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

Picture of distance formula and graph

Video Tutorial on the Distance Formula

Practice Problems

What is the distance between the the points $$(0,0)$$ and $$(6,8)$$ plotted on the graph?

The Distance Formula Distance Formula picture and problem
Step 1

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 $

Step 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 $$ (8)^2 $$

Problem 2

What is the distance between the points $$ ( 2, 4) $$ and $$ (26, 9)$$ ? Round your answer to the nearest tenth

Step 1

Graph the two points

Step 2

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 $
Step 3

Solve for the Hyptenuse

$ a^2 + b^2 = \red c^2 \\ 5^2 + 24^2 = \red c^2 \\ \color{green}{ \sqrt{25 + 576 }=c } \\ \sqrt 601 = \red {c} \\ \boxed { \red c = \sqrt 601 = 24.5 } $

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 } $

Problem 3

What is the distance between the points $$ ( 4, 6 ) $$ and $$ (28, 13 )$$ ? Round your answer to the nearest tenth

Step 1

Graph the two points

Step 2

Plug in the side lengths into the Pythagoren Theorem

$ a^2 + b^2 = c^2 \\ 24^2 + 7^2 = \red c^2 $
Step 3

Solve for the Hyptenuse

$ a^2 + b^2 = \red c^2 \\ 24^2 + 7^2 = \red c^2 \\ \color{green}{ \sqrt{576+ 49}=c } \\ \sqrt{ 625} = \red c \\ \boxed { \red c = 25 } $

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 } $

Problem 4

The point $$(4,8)$$ lies on a circle centered at $$ (12, 14)$$. What is the radius of this circle ? Round your answer to the nearest tenth

Stuck? Click here for a big hint
Step 1

Graph the two points

Step 2

Plug in the side lengths into the Pythagoren Theorem

$ a^2 + b^2 = c^2 \\ 6^2 + 8^2 = \red c^2 $
Step 3

Solve for the Hyptenuse

$ a^2 + b^2 = \red c^2 \\ 6^2 + 8^2 = \red c^2 \\ \color{green}{ \sqrt{36 + 64}=c } \\ \sqrt{ 100 } = \red c \\ \boxed { \red c = 10 } $

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 } $