IGCSE Math RevisionQuestion 1
Find the distance between the points \((3, -1)\) and \((-5, 5)\).
Show the mark scheme and worked solution
- M1 √((3 − (−5))² + (−1 − 5)²)
- A1 10
- Differences: 3 − (−5) = 8 and −1 − 5 = −6.
- Distance = √(8² + 6²) = √100 = 10.
Answer: \(10\)
IGCSE Additional Maths · Cambridge 0606 · Edexcel 4PM1
Coordinate geometry: notes and questions
Revision for straight lines, gradients, parallel and perpendicular lines, midpoints and lengths, perpendicular bisectors, dividing a line in a ratio (4PM1), and circles (0606). Both boards cover straight lines. Equations of circles are Cambridge 0606 only; dividing a line in a given ratio is in Edexcel 4PM1.
- Cambridge 0606: section 7, 8
- Edexcel 4PM1: section 8
- 14 questions
Key points
- Gradient = (y₂ − y₁)/(x₂ − x₁). Parallel lines have equal gradients; perpendicular gradients multiply to −1.
- Use y − y₁ = m(x − x₁) for a line through a known point.
- The perpendicular bisector of AB passes through the midpoint of AB with gradient −1/m(AB).
- The point dividing AB in the ratio m : n is ((nx₁ + mx₂)/(m + n), (ny₁ + my₂)/(m + n)).
- A circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². Complete the square to read the centre and radius from the expanded form.
- A tangent to a circle is perpendicular to the radius at the point of contact.
Watch out: When the question asks for ax + by + c = 0 with integers, clear every fraction and check the signs at the end.
Worked example
The points \(A\) and \(B\) have coordinates \((-2, 5)\) and \((4, 1)\).
Find the equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.
- B1 midpoint (1, 3)
- B1 gradient of AB −2/3
- M1 perpendicular gradient 3/2 with the midpoint
- A1 3x − 2y + 3 = 0
- Midpoint of AB: ((−2 + 4)/2, (5 + 1)/2) = (1, 3).
- Gradient of AB: (1 − 5)/(4 − (−2)) = −4/6 = −2/3.
- Perpendicular gradient: 3/2. Line: y − 3 = (3/2)(x − 1).
- 2y − 6 = 3x − 3, so 3x − 2y + 3 = 0.
Answer: \(3x - 2y + 3 = 0\)
Coordinate geometry questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 2
Find the equation of the line through \((2, -3)\) that is parallel to the line \(4x + 2y = 7\).
Show the mark scheme and worked solution
- B1 gradient −2
- M1 y + 3 = −2(x − 2)
- A1 y = −2x + 1
- 4x + 2y = 7 gives y = −2x + 3.5, gradient −2.
- Parallel line: y − (−3) = −2(x − 2).
- y = −2x + 4 − 3 = −2x + 1.
Answer: \(y = -2x + 1\)
Question 3
The point \(P(5, 3)\) lies on the circle \((x - 2)^2 + (y + 1)^2 = 25\).
Find the equation of the tangent to the circle at \(P\), giving your answer in the form \(ax + by = c\).
Show the mark scheme and worked solution
- B1 gradient of radius 4/3
- M1 tangent gradient −3/4 through (5, 3)
- A1 3x + 4y = 27
- The centre is (2, −1). Gradient of the radius to P: (3 − (−1))/(5 − 2) = 4/3.
- A tangent is perpendicular to the radius: gradient −3/4.
- y − 3 = −¾(x − 5), so 4y − 12 = −3x + 15 and 3x + 4y = 27.
Answer: \(3x + 4y = 27\)
Question 4
Find the values of \(k\) for which the line \(y = x + k\) is a tangent to the circle \(x^2 + y^2 = 8\).
Show the mark scheme and worked solution
- M1 2x² + 2kx + k² − 8 = 0
- M1 discriminant zero
- A1 k = 4 or k = −4
- x² + (x + k)² = 8, so 2x² + 2kx + k² − 8 = 0.
- Tangent: (2k)² − 4 × 2 × (k² − 8) = 0, so 4k² − 8k² + 64 = 0.
- k² = 16, so k = ±4.
Answer: \(k = 4\) or \(k = -4\)
9 more coordinate geometry questions
Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.
Next steps
- Cambridge 0606: papers, syllabus map and official past papers
- Edexcel 4PM1: papers, syllabus map and official past papers
- Going on to A Level? Coordinate geometry continues in A Level coordinate geometry (A Level Math Revision). See also from Additional Maths to A Level.