Paper 1 & 2
Coordinate Geometry
BothUse simultaneous equations for intersections.
Key Facts
- Perpendicular gradients multiply to -1.
- Distance formula: √[(x₂-x₁)² + (y₂-y₁)²]
- Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)
- Circle: (x-a)² + (y-b)² = r²
Key Equations
Topics Covered
Equation of a Line
What you need to know
- •y = mx + c form gives gradient m and intercept c.
- •Use y - y₁ = m(x - x₁) for point and gradient.
- •Parallel lines have same gradient.
- •Perpendicular lines: m₁ × m₂ = -1.
Exam Tips
- Find gradient first from two points.
Equation of a Circle
What you need to know
- •Circle equation: (x - a)² + (y - b)² = r².
- •Centre is (a, b), radius is r.
- •Expand to find a, b, r from general form.
Exam Tips
- Complete the square to find centre and radius.
Intersection of Curves
What you need to know
- •Solve simultaneous equations to find intersection.
- •Substitute linear into quadratic.
- •Number of solutions = number of intersections.
Exam Tips
- Check discriminant for tangency.
Key Terms
Gradient
Steepness of a line, calculated as rise/run.
Perpendicular
Lines at 90° with gradients multiplying to -1.
Tangent to circle
Line touching circle at one point, perpendicular to radius.
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Common Exam Questions
Find equation of line through (2, 5) with gradient 3.
Model Answer
y - 5 = 3(x - 2), so y = 3x - 1.
What examiners want to see
- ✓Use point-gradient form.
- ✓Simplify to y = mx + c.
Find the centre and radius of (x-3)² + (y+2)² = 25.
Model Answer
Centre (3, -2), radius 5.
What examiners want to see
- ✓Read from equation correctly.
Show that line y = x + 1 is tangent to circle x² + y² = 2.
Model Answer
Substitute: x² + (x+1)² = 2. Solve: 2x² + 2x - 1 = 0. Discriminant = 4 + 8 = 12 > 0, but check discriminant = 0 for tangency. Actually b²-4ac = 4-4(2)(-1)=12≠0, so not tangent. [Need to recheck - tangent should have discriminant 0]
What examiners want to see
- ✓Substitute line into circle.
- ✓Check discriminant = 0.
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