How to Solve: Coordinate Geometry (Equations of Lines, Circles, Parametric Equations)

How to Solve: Coordinate Geometry (Equations of Lines, Circles, Parametric Equations)

Exam Context: GCSE (Higher Tier) / A-Level Maths (Pure) Score Impact: 10–15% of your exam paper. Master this, and you’ll breeze through graph questions, save time, and pick up easy marks on multi-step problems.

Introduction

"Imagine you’re an air traffic controller tracking two planes on a radar screen. One plane’s path is a straight line, the other’s a perfect circle. To predict if they’ll collide, you need to write their equations—and that’s exactly what this topic teaches. Miss it, and you’re leaving 15 marks on the table."

What You Need To Know First

1. Plotting points on a grid – You must know how to read (x, y) coordinates.
2. Basic algebra – Rearranging equations, solving for y, and substituting values.
3. Pythagoras’ theorem – Used in distance formulas and circle equations.

Key Vocabulary

Formulas To Know

1. Equation of a Straight Line

Formula: y = mx + c - m = gradient (steepness) - c = y-intercept MEMORISE THIS

Alternative Formula (Point-Gradient Form): y – y₁ = m(x – x₁) - (x₁, y₁) = a point on the line - m = gradient MEMORISE THIS

Gradient Formula (Two Points): m = (y₂ – y₁) / (x₂ – x₁) - (x₁, y₁) and (x₂, y₂) = two points on the line MEMORISE THIS

2. Equation of a Circle

Standard Form: (x – a)² + (y – b)² = r² - (a, b) = centre of the circle - r = radius MEMORISE THIS

Expanded Form (Given on Exam Sheet): x² + y² + 2gx + 2fy + c = 0 - Centre = (–g, –f) - Radius = √(g² + f² – c) Given on exam sheet (but know how to use it!)

3. Parametric Equations

General Form: x = f(t), y = g(t) - t = parameter (usually time or angle) - Eliminate t to find the Cartesian equation. MEMORISE THIS

Example: x = 3t + 1, y = 2t – 4 → Eliminate t to get y = (2/3)x – 14/3

4. Distance Between Two Points

Formula: d = √[(x₂ – x₁)² + (y₂ – y₁)²] MEMORISE THIS

5. Midpoint of a Line Segment

Formula: Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) MEMORISE THIS

Step-by-Step Method

How to Find the Equation of a Line

Given: - Two points, OR - One point and a gradient.

Steps: 1. Find the gradient (m) using m = (y₂ – y₁)/(x₂ – x₁). 2. Pick one point (x₁, y₁). 3. Plug into y – y₁ = m(x – x₁). 4. Rearrange to y = mx + c (if needed).

Example: Find the equation of the line through (2, 3) and (4, 7). 1. m = (7 – 3)/(4 – 2) = 4/2 = 2 2. Use point (2, 3). 3. y – 3 = 2(x – 2) 4. y = 2x – 4 + 3 → y = 2x – 1

How to Find the Equation of a Circle

Given: - Centre (a, b) and radius r, OR - Three points on the circle.

Steps (Centre & Radius Given): 1. Write down (a, b) and r. 2. Plug into (x – a)² + (y – b)² = r². 3. Expand if needed (but keep it neat).

Example: Circle with centre (3, -2) and radius 5. 1. (a, b) = (3, -2), r = 5 2. (x – 3)² + (y + 2)² = 25

Steps (Three Points Given): 1. Find the perpendicular bisectors of two chords. 2. Find their intersection (this is the centre). 3. Find the radius (distance from centre to any point). 4. Write the equation.

How to Work with Parametric Equations

Given: x = f(t), y = g(t)

Steps: 1. Express t in terms of x (or y). 2. Substitute into the other equation. 3. Simplify to get y = mx + c (or circle equation).

Example: x = 2t + 1, y = t² – 3 1. t = (x – 1)/2 2. y = [(x – 1)/2]² – 3 3. y = (x² – 2x + 1)/4 – 3 → y = (1/4)x² – (1/2)x – 11/4

Worked Examples

Example 1 – Basic (Line Equation)

Question: Find the equation of the line passing through (1, 4) and (3, 10).

Solution: 1. m = (10 – 4)/(3 – 1) = 6/2 = 3 2. Use point (1, 4). 3. y – 4 = 3(x – 1) 4. y = 3x – 3 + 4 → y = 3x + 1

What we did and why: - Found the gradient first (step 1). - Used point-gradient form (step 3) to avoid mistakes with c. - Rearranged to y = mx + c for the final answer.

Example 2 – Medium (Circle Equation)

Question: A circle has centre (2, -1) and passes through (5, 3). Find its equation.

Solution: 1. Find radius: r = √[(5 – 2)² + (3 – (-1))²] = √(9 + 16) = 5 2. Plug into (x – a)² + (y – b)² = r². 3. (x – 2)² + (y + 1)² = 25

What we did and why: - Used the distance formula to find r (step 1). - Kept the equation in standard form (step 3) for full marks.

Example 3 – Exam-Style (Parametric to Cartesian)

Question: A curve is given by x = 3t – 2, y = 2t² + 1. Find its Cartesian equation.

Solution: 1. From x = 3t – 2, t = (x + 2)/3. 2. Substitute into y = 2t² + 1:
y = 2[(x + 2)/3]² + 1 3. y = 2(x² + 4x + 4)/9 + 1 4. y = (2x² + 8x + 8)/9 + 9/9 5. y = (2x² + 8x + 17)/9

What we did and why: - Eliminated t by substitution (steps 1–2). - Simplified carefully (steps 3–5) to avoid sign errors.

Common Mistakes

Exam Traps

1-Minute Recap

"Right, listen up—this is your last-minute cheat sheet. For lines, remember: gradient first, then point-gradient form. For circles, centre and radius—always write (x – a)² + (y – b)² = r². If you see x = t + 1, y = 2t, eliminate t to get y = 2x – 2. Watch out for perpendicular lines (m × m = -1) and expanded circle equations (complete the square!). And if you’re stuck, draw a quick sketch—it’ll save you every time. Now go smash that exam!