Coordinate Geometry – Algebra Meets Geometry

Coordinate geometry (also called analytical geometry) brings together algebra and geometry. By placing shapes on a numbered grid, we can use equations to describe, analyse, and solve geometric problems precisely.

The system is often called Cartesian geometry after René Descartes, whose Latinised name was Cartesius, but the French lawyer and mathematician Pierre de Fermat developed strikingly similar ideas at almost exactly the same time in the 1630s, in a manuscript that was not published until after his death; today the two are usually credited jointly as the founders of coordinate geometry. The (x, y) coordinate idea turned out to be one of the most quietly powerful tools ever invented in mathematics, since it lets purely algebraic equations be drawn as pictures, and purely visual shapes be manipulated with algebra instead of a ruler and compass. Every digital image relies on the same idea today: a computer screen is really just a giant coordinate grid of pixels, and every line, circle, or curve drawn in design software is calculated using exactly the formulas on this page.

The Coordinate System

Every point in a plane is described by an ordered pair (x, y). The x-axis runs horizontally; the y-axis runs vertically. They meet at the origin (0, 0). Positive x goes right, positive y goes up.

Key Formulas

Formula	Expression	Use
Distance	d = √[(x₂−x₁)² + (y₂−y₁)²]	Length of a segment between two points
Midpoint	M = ((x₁+x₂)/2, (y₁+y₂)/2)	Centre point of a segment
Gradient (slope)	m = (y₂−y₁)/(x₂−x₁)	Steepness of a line
Equation of line	y = mx + c	Describes any straight line
Parallel lines	Same gradient m	Never meet
Perpendicular lines	m₁ × m₂ = −1	Meet at 90°

Where the Distance Formula Comes From

The distance formula isn't a separate rule to memorise — it's just Pythagoras' theorem in disguise. To find the distance between two points, build a right-angled triangle between them: one leg going straight across (the horizontal distance), one leg going straight up (the vertical distance), and the segment you want is the hypotenuse.

The horizontal leg has length |8 − 2| = 6 (the difference in x-values, x₂ − x₁). The vertical leg has length |11 − 3| = 8 (the difference in y-values, y₂ − y₁). By Pythagoras' theorem, the hypotenuse d satisfies d² = 6² + 8² = 36 + 64 = 100, so d = √100 = 10. Writing that in general symbols — using (x₂−x₁) and (y₂−y₁) as the two legs, for any pair of points — gives exactly the distance formula: d = √[(x₂−x₁)² + (y₂−y₁)²].

Worked Examples

Find the distance between A(2, 3) and B(8, 11).

d = √[(8−2)² + (11−3)²] = √[36 + 64] = √100 = 10.

Find the midpoint of P(1, 5) and Q(7, 9).

M = ((1+7)/2, (5+9)/2) = (4, 7). Midpoint: (4, 7).

Find the equation of the line through (0, 3) with gradient 2.

y = mx + c. m = 2, c = 3 (y-intercept). Equation: y = 2x + 3.

Find the equation of the line through (2, 5) and (4, 9).

Gradient: m = (9−5)/(4−2) = 2. Using y−5 = 2(x−2): y = 2x + 1. Equation: y = 2x + 1.

Key Takeaways

Distance formula: d = √[(Δx)² + (Δy)²] — based on Pythagoras.
Midpoint = average of the x-coordinates and average of the y-coordinates.
Gradient = rise / run = (y₂−y₁)/(x₂−x₁).
Perpendicular gradients multiply to −1.

Practice: Equations & Gradients

Find the Equation of a Line