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1. Distance Formula

See also:

The Cartesian Plane

The cartesian plane was named after Rene Descartes.

See more about Descartes in Functions and Graphs.

We have a right-angled triangle with hypotenuse length c, as shown:

Recall Pythagoras' Theorem, which tells us the length of the longest side (the hypotenuse) of a right triangle:

`c=sqrt(a^2+b^2)`

We use this to find the distance between any two points (x1, y1) and (x2, y2) on the cartesian (x-y) plane:

x y
A (x1, y1)
B (x2, y1)
C (x2, y2)
distance = d

The point B (x2, y1) is at the right angle. We can see that:

  • The distance between the points A(x1, y1) and B(x2, y1) is simply x2x1 and
  • The distance between the points C(x2, y2) and B(x2, y1) is simply y2y1.
x y
A (x1, y1)
B (x2, y1)
C (x2, y2)
x2 − x1
y2 − y1
distance = d

Distance from (x1, y1) to (x2, y2).

Using Pythagoras' Theorem we can develop a formula for the distance d.

Continues below

Distance Formula

The distance between (x1, y1) and (x2, y2) is given by:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2`

Note: Don't worry about which point you choose for (x1, y1) (it can be the first or second point given), because the answer works out the same.

Interactive Graph - Distance Formula

You can explore the concept of distance formula in the following interactive graph (it's not a fixed image).

Drag either point A (x1, y1) or point C (x2, y2) to investigate how the distance formula works.

Length AB = x2x1

Length BC = y2y1

Length

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Example 1

Find the distance between the points (3, −4) and (5, 7).

Answer

Here, x1 = 3 and y1 = −4; x2 = 5 and y2 = 7

So the distance is given by:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((5-3)^2+(7-(-4))^2)`

`=sqrt(4+121)`

`=11.18`

Example 2

Find the distance between the points (3, −1) and (−2, 5).

Answer

This time, x1 = 3 and y1 = −1; x2 = −2 and y2 = 5

So the distance is given by:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-2-3)^2+(5-(-1))^2)`

`=sqrt(25+36)`

`=sqrt61`

`=7.8102`

Example 3

What is the distance between (−1, 3) and (−8, −4)?

Answer

In this example, x1 = −1 and y1 = 3; x2 = −8 and y2 = −4

So the distance is given by:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-8-(-1))^2+(-4-3)^2)`

`=sqrt(49+49)`

`=sqrt98`

`=9.899`

Example 4

Find k if the distance between (k,0) and (0, 2k) is 10 units.

Answer

This is the situation:

Graph solution - straight line

Applying the distance formula, we have:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((2k-0)^2+(0-k)^2)`

`=sqrt(4k^2+k^2)`

`=sqrt(5k^2)`

Now `sqrt(5k^2)=10` so `5k^2=100`, giving:

k2 = 20

so

`k=+-sqrt(20)~~+-4.472`

We obtained 2 solutions, so there are 2 possible outcomes, as follows:

Graph solutions - 2 straight line

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