Pure 1: Coordinate Geometry

Pure 1: Coordinate Geometry — CIE A Level Mathematics 9709

1. Straight Lines

Key formulae:

| Concept | Formula |

|---|---|

| Gradient | m = (y₂ − y₁)/(x₂ − x₁) |

| Distance | d = √[(x₂−x₁)² + (y₂−y₁)²] |

| Midpoint | M = ((x₁+x₂)/2, (y₁+y₂)/2) |

| Line through (x₁,y₁) with gradient m | y − y₁ = m(x − x₁) |

| General form | ax + by + c = 0 |

Parallel lines: Same gradient m₁ = m₂.

Perpendicular lines: m₁ × m₂ = −1 (product of gradients = −1).

Example: Line L has equation 2x + 3y = 6. Find the equation of the perpendicular through (1, 2).

• Gradient of L: rearrange → y = −(2/3)x + 2, so m₁ = −2/3
• Perpendicular gradient: m₂ = 3/2
• Line: y − 2 = (3/2)(x − 1) → 2y = 3x + 1 or 3x − 2y + 1 = 0

Perpendicular bisector: Passes through midpoint of AB, perpendicular to AB.

2. Circles

Standard form: (x − a)² + (y − b)² = r² — centre (a, b), radius r.

General form: x² + y² + 2gx + 2fy + c = 0 — centre (−g, −f), radius √(g² + f² − c).

Completing the square to find centre and radius:

x² + y² − 4x + 6y + 4 = 0

(x−2)² − 4 + (y+3)² − 9 + 4 = 0

(x−2)² + (y+3)² = 9 → centre (2, −3), radius 3.

Key circle theorems (used in 9709):

• The angle in a semicircle is 90° — useful for showing a line is a diameter
• A tangent to a circle is perpendicular to the radius at the point of contact
• The perpendicular from the centre bisects a chord

Tangent to a circle at point P(x₁, y₁):

If circle is (x−a)² + (y−b)² = r², the radius to P has gradient (y₁−b)/(x₁−a).

Tangent gradient = −(x₁−a)/(y₁−b). Then use point-slope form.

3. Intersection of Line and Circle

Substitute the line equation into the circle equation → quadratic in x (or y).

• Δ > 0: two intersection points
• Δ = 0: line is tangent to circle
• Δ < 0: line misses circle

Example: Find where y = x + 1 meets x² + y² = 5.

Substitute: x² + (x+1)² = 5 → 2x² + 2x − 4 = 0 → x² + x − 2 = 0 → (x+2)(x−1) = 0.

x = −2, y = −1 and x = 1, y = 2. Points: (−2, −1) and (1, 2).

4. Parametric Equations

A curve can be defined by x = f(t) and y = g(t) where t is the parameter.

To find the Cartesian equation: eliminate t.

Example: x = 2t, y = t² − 1. Find the Cartesian equation.

t = x/2 → y = (x/2)² − 1 = x²/4 − 1 (parabola).

Finding dy/dx for parametric curves:

dy/dx = (dy/dt) ÷ (dx/dt)

Example: x = t², y = 2t. dy/dt = 2, dx/dt = 2t. So dy/dx = 2/(2t) = **1/t**.

5. Locus Problems

A locus is the set of all points satisfying a given condition.

Example: Find the locus of point P(x, y) equidistant from A(1, 0) and B(5, 4).

|PA|² = |PB|²

(x−1)² + y² = (x−5)² + (y−4)²

x²−2x+1 + y² = x²−10x+25 + y²−8y+16

8x + 8y = 40 → x + y = 5 (perpendicular bisector of AB).