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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 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

FormulaExpressionUse
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°

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 Questions

  1. Find the distance between C(3, −1) and D(7, 2).
  2. Find the midpoint of (5, −4) and (1, 8).
  3. A line has gradient 3 and passes through (0, −2). Write its equation.
  4. Find the gradient of the perpendicular to the line y = 4x + 1.
  5. Write the equation of the line through (3, 7) perpendicular to y = 3x − 5.
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