The Distance Formula Explorer An interactive guide to understanding the distance between two points in a plane

The Distance Formula Explorer

An interactive guide to understanding the distance between two points in a plane.

The Magic Formula

To find the distance \( d \) between any two points, \( P_1(x_1, y_1) \) and \( P_2(x_2, y_2) \), we use the following formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Looks familiar? It's really just the Pythagorean theorem in disguise! Let's find out how.

Bring it to Life!

Click or tap anywhere on the graph to place two points. You can also drag the points around to see how the distance and calculations change in real-time.

Live Calculation

Point A (x₁, y₁): (-4, -2)

Point B (x₂, y₂): (5, 6)

1. Find the horizontal change \( (\Delta x) \):

\( x_2 - x_1 = 5 - (-4) = 9 \)

2. Find the vertical change \( (\Delta y) \):

\( y_2 - y_1 = 6 - (-2) = 8 \)

3. Apply the formula:

\( d = \sqrt{9^2 + 8^2} \)

\( d = \sqrt{81 + 64} = \sqrt{145} \)

Final Distance (d):

≈ 12.04

Derivation: It's All About Triangles!

The distance formula is a direct application of the Pythagorean Theorem, which states that for a right-angled triangle with sides 'a' and 'b' and hypotenuse 'c', we have:

$$ a^2 + b^2 = c^2 $$

Now, let's see how this relates to our coordinate plane.

1. Imagine two points, \( P_1(x_1, y_1) \) and \( P_2(x_2, y_2) \). The line segment connecting them is the distance \( d \) we want to find.
2. We can form a right-angled triangle where this line segment is the hypotenuse.
3. The length of the horizontal side ('a') is the difference between the x-coordinates: \( |x_2 - x_1| \).
4. The length of the vertical side ('b') is the difference between the y-coordinates: \( |y_2 - y_1| \).
5. Plugging these into the Pythagorean theorem (\( c = \sqrt{a^2 + b^2} \)), we get our distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

The interactive graph above visually constructs this exact triangle for you!

Solved Examples

Let's walk through the examples from your textbook images.

Example 1: Points P(4, 6) and Q(6, 8)

Here \( (x_1, y_1) = (4, 6) \) and \( (x_2, y_2) = (6, 8) \).

$$ d = \sqrt{(6 - 4)^2 + (8 - 6)^2} $$ $$ d = \sqrt{(2)^2 + (2)^2} $$ $$ d = \sqrt{4 + 4} = \sqrt{8} $$ $$ d = 2\sqrt{2} \approx 2.83 $$

Example 2: Points P(6, 4) and Q(-5, -3)

Here \( (x_1, y_1) = (6, 4) \) and \( (x_2, y_2) = (-5, -3) \). Be careful with the negative signs!

$$ d = \sqrt{(-5 - 6)^2 + (-3 - 4)^2} $$ $$ d = \sqrt{(-11)^2 + (-7)^2} $$ $$ d = \sqrt{121 + 49} = \sqrt{170} $$ $$ d \approx 13.04 $$

Special Case: Distance from Origin O(0, 0)

To find the distance of any point \( P(x, y) \) from the origin, the formula simplifies because \(x_1=0\) and \(y_1=0\).

$$ d = \sqrt{(x - 0)^2 + (y - 0)^2} $$ $$ d = \sqrt{x^2 + y^2} $$