Distance Between Two Points

Statement

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a coordinate plane can be calculated using Pythagoras’ theorem. The formula is:

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

This gives the straight-line (Euclidean) distance between the two points.

Why it’s true

• The horizontal difference between the two points is \(|x_2 - x_1|\).
• The vertical difference is \(|y_2 - y_1|\).
• Together, these form the legs of a right-angled triangle, with the distance \(d\) as the hypotenuse.
• By Pythagoras’ theorem, hypotenuse squared equals sum of squares of the other two sides.

Recipe (how to use it)

1. Label the two points clearly as \((x_1, y_1)\) and \((x_2, y_2)\).
2. Subtract the x-coordinates and y-coordinates to find the differences.
3. Square both differences.
4. Add the squares together and take the square root.

Spotting it

This formula applies whenever you are asked for the distance between two points on a coordinate grid, or the length of a line segment given its endpoints.

Common pairings

• Midpoint formula: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
• Gradient formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
• Equation of a circle: \((x-h)^2 + (y-k)^2 = r^2\), which is based on distance from centre to point.

Mini examples

1. Points: \((0,0)\) and \((3,4)\). Answer: \(d = \sqrt{3^2 + 4^2} = 5\).
2. Points: \((1,2)\) and \((4,6)\). Answer: \(d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5\).

Pitfalls

• Forgetting to square the differences before adding.
• Dropping negative signs incorrectly (squaring removes them anyway).
• Confusing distance formula with gradient formula.

Exam strategy

• Write coordinates clearly and substitute carefully.
• Check arithmetic when squaring and adding values.
• Decide if the answer should be left as a surd or given as a decimal (depending on the question).

Summary

The distance formula is a direct application of Pythagoras’ theorem in coordinate geometry. By squaring the horizontal and vertical differences and adding, then square rooting, you get the straight-line distance between two points.