Coordinate geometry

OCR J560 weaves coordinate geometry through both Foundation and Higher: midpoints, distances, gradients, equations of lines, and the geometric properties of polygons placed on axes.

Distance between two points

Pythagoras' theorem in disguise. For (x₁, y₁) and (x₂, y₂): d = √((x₂ − x₁)² + (y₂ − y₁)²).

E.g. distance from (1, 2) to (4, 6): d = √(3² + 4²) = √25 = 5.

Midpoint

M = ((x₁ + x₂)/2, (y₁ + y₂)/2). The averages of the coordinates.

E.g. midpoint of (1, 2) and (4, 6) is (2.5, 4).

Gradient

m = (y₂ − y₁) / (x₂ − x₁).

Equation of a line through two points

1. Compute the gradient.
2. Use y − y₁ = m(x − x₁).

E.g. through (1, 2) and (4, 6): m = 4/3. y − 2 = (4/3)(x − 1) → y = (4/3)x + 2/3.

Geometric proofs on axes

OCR Higher problems often ask: prove ABCD is a parallelogram / rectangle / rhombus / square.

Parallelogram: opposite sides are parallel (equal gradients) and equal in length. Rectangle: parallelogram with all angles 90° (adjacent sides perpendicular: gradients multiply to −1). Rhombus: parallelogram with all four sides equal. Square: rhombus with right angles.

For an isosceles triangle, show two sides are equal in length.

Equation of a perpendicular bisector

The perpendicular bisector of segment AB:

1. Midpoint M of AB.
2. Gradient of AB: m.
3. Perpendicular gradient: −1/m.
4. Equation: y − M_y = −1/m · (x − M_x).

This is the locus of all points equidistant from A and B — useful for circle centres and triangle circumcentres.

OCR mark scheme conventions

• M1 for stating an appropriate formula (distance, midpoint, gradient).
• M1 for substituting the correct values.
• A1 for the answer (cao or to the required precision).
• "Show that" demands the working — answer alone scores zero.