Distance Between Two Points

Learn how to find the distance between two points using the distance formula derived from the Pythagorean theorem.

1. Need for Distance Formula

In a Cartesian plane, points are located using coordinates. To find how far two points are from each other, we need a method that works for all positions—same quadrant or different quadrants. The distance formula helps us measure the straight-line distance between any two points.

1.1. What We Already Know

The shortest distance between two points is always a straight line, and in many cases, we can think of this line as the hypotenuse of a right-angled triangle.

2. Derivation Using Pythagoras Theorem

The distance formula is taken directly from the Pythagorean theorem. If we draw a right triangle between two points, the horizontal and vertical distances form the two perpendicular sides.

2.1. Right Triangle Idea

Consider two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\). If we join them, we can form a right-angled triangle by drawing horizontal and vertical lines through the points.

2.2. Using Pythagoras

The horizontal distance = \(|x_2 - x_1|\).
The vertical distance = \(|y_2 - y_1|\).

According to the Pythagorean theorem:

\( \text{Distance}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \)

2.2.1. Taking Square Root

Taking the square root gives the final formula.

3. Distance Formula

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) on the Cartesian plane is:

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

3.1. Key Points

  • The order of coordinates does not matter because the squares make the values positive.
  • The formula works for all quadrants.
  • The distance is always positive.

4. Examples

These examples show how to use the distance formula in different situations.

4.1. Example 1: Same Quadrant

Find the distance between \((2, 3)\) and \((6, 9)\).

Using the formula:

\( d = \sqrt{(6 - 2)^2 + (9 - 3)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \)

4.2. Example 2: Different Quadrants

Find the distance between \((-4, 2)\) and \((3, -1)\).

\( d = \sqrt{(3 + 4)^2 + (-1 - 2)^2} = \sqrt{7^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} \)

4.3. Example 3: Vertical or Horizontal Points

Between \((5, 2)\) and \((5, -4)\):

The x-coordinates are the same, so the distance is:

\( d = |2 - (-4)| = 6 \)