# 1. Distance Formula

We use the Distance Forumala to find the distance between any two points (x1,y1) and (x2,y2) on a cartesian plane.

### The Cartesian Plane

The cartesian plane was named after Rene Descartes.

See more about Descartes in Functions and Graphs.

Let's start with 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:

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.
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.

## 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. As you drag the points the graph will automatically calculate the distance.

Length AB = x2x1

Length BC = y2y1

Length

### Example 1

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

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).

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)?

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.

This is the situation: 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: Search IntMath

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