# Distance Between 2 Points

Here is how to calculate the distance between two points when you know their coordinates:

Let us call the two points A and B

We can run lines down from A, and along from B, to make a Right Angled Triangle.

And with a little help from Pythagoras we know that:

a^{2} + b^{2} = c^{2}

Now label the coordinates of points A and B.

x_{A} means the x-coordinate of point A

y_{A} means the y-coordinate of point A

The horizontal distance **a** is **(x _{A} − x_{B})**

The vertical distance **b** is **(y _{A} − y_{B})**

Now we can solve for c (the distance between the points):

^{2}= a

^{2}+ b

^{2}

^{2}= (x

_{A}− x

_{B})

^{2}+ (y

_{A}− y

_{B})

^{2}

## Examples

### Example 1

Fill in the values: | ||

### Example 2

It doesn't matter what order the points are in, because squaring removes any negatives:

Fill in the values: | ||

### Example 3

And here is another example with some negative coordinates ... it all still works:

Fill in the values: | ||

*(Note √136 can be further simplified to 2√34 if you want)*

## Try It Yourself

Drag the points:

## Three or More Dimensions

It works perfectly well in 3 (or more!) dimensions.

Square the difference for each axis, then sum them up and take the square root:

Distance = √[ (x_{A} − x_{B})^{2} + (y_{A} − y_{B})^{2} + (z_{A} − z_{B})^{2} ]

### Example: the distance between the two points (8,2,6) and (3,5,7) is:

= √[ (8−3)^{2} + (2−5)^{2} + (6−7)^{2} ] |

= √[ 5^{2} + (−3)^{2} + (−1)^{2} ] |

= √( 25 + 9 + 1 ) |

= √35 |

Which is about 5.9 |