Cambridge International AS & A Level Mathematics 9709: Coordinate Geometry (Pure Mathematics 1) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 1.3, Coordinate geometry, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The content is part of Pure Mathematics 1 and is examined on Paper 1, which is compulsory for AS Level and A Level. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give exact answers (surds and fractions) unless a question asks otherwise.

Related: the coordinate geometry study guide, the revision notes, the quadratics practice questions, the A Level Mathematics hub, the printable 9709 checklist and the free AS Level diagnostic.

Questions

1. The points A and B have coordinates (−3, 4) and (5, −2). Find the length of AB and the coordinates of the midpoint of AB. [3]

2. The line L has equation 3x − 2y + 7 = 0. The point P has coordinates (7, 1).

(a) Find the equation of the line through P perpendicular to L, giving your answer in the form ax + by + c = 0 where a, b and c are integers. [4] (b) Find the coordinates of the point N where this line meets L. [2] (c) Hence find the shortest distance from P to L, giving your answer as an exact surd. [1]

3. The points A(2, 7) and B(8, −1) are given.

(a) Find the equation of the perpendicular bisector of AB. [4] (b) The perpendicular bisector meets the x-axis at C. Find the area of triangle ABC. [3]

4. A circle has equation x² + y² + 8x − 10y + 16 = 0.

(a) Find the coordinates of the centre and the radius of the circle. [3] (b) Show that the point P(−1, 1) lies on the circle, and find the equation of the tangent to the circle at P. [4]

5. The points A(−1, −2) and B(7, 4) are the ends of a diameter of a circle.

(a) Find the equation of the circle. [3] (b) Verify that the point C(6, 5) lies on the circle. [1] (c) Show that AC is perpendicular to BC, and state the property of circles that this illustrates. [2]

6. The circle x² + y² − 2x + 4y − 20 = 0 and the line x + y = 4 meet at the points P and Q.

(a) Find the coordinates of P and Q. [4] (b) Show that the perpendicular bisector of PQ passes through the centre of the circle. [3]

7. The line y = x + k and the curve y = x² − 5x + 7 are given, where k is a constant.

(a) Find the value of k for which the line is a tangent to the curve, and find the coordinates of the point where it touches. [4] (b) State the set of values of k for which the line does not meet the curve. [1]

8. Find the two values of c for which the line y = 2x + c is a tangent to the circle x² + y² = 20. For each value, find the coordinates of the point of contact. [6]

9. A circle passes through the points A(1, 4) and B(7, 2). Its centre lies on the line y = x − 3.

(a) Find the equation of the perpendicular bisector of AB. [3] (b) Find the coordinates of the centre, and hence the equation of the circle. [4] (c) Find the equation of the tangent to the circle at A, in the form ax + by + c = 0. [3] (d) The tangent at A meets the x-axis at T. Find the exact length of TA. [2]

10. The points A(0, 6) and B(8, 2) are given. The point C lies on the x-axis and angle ACB = 90°. Find the possible coordinates of C. [5]

11. The line y = mx passes through the origin. Find the set of values of m for which the line meets the circle (x − 5)² + y² = 9 at two distinct points. [5]

12. ABCD is a rhombus. A is the point (1, 2) and C is the point (9, 6). B lies on the y-axis.

(a) Find the equation of the diagonal BD. [3] (b) Find the coordinates of B and D. [3] (c) Find the area of the rhombus. [3]

Answers

1. AB = √(8² + 6²) [1] = 10 [1]. Midpoint (1, 1) [1] Examiner insight: the method mark for the length needs both differences squared inside one square root; √8² + √6² earns nothing.

2. (a) Gradient of L = 3/2 [1]. Perpendicular gradient = −2/3 [1]. y − 1 = −(2/3)(x − 7) [1], so 2x + 3y − 17 = 0 [1] (b) Solve 3x − 2y + 7 = 0 and 2x + 3y − 17 = 0 simultaneously [1]: N(1, 5) [1] (c) PN = √(6² + 4²) = √52 = 2√13 [1] Examiner insight: the final mark in (a) is for the requested form with integers; a correct line left as y = −(2/3)x + 17/3 loses it.

3. (a) Midpoint (5, 3) [1]. Gradient AB = −8/6 = −4/3 [1], so the perpendicular gradient is 3/4 [1]. y − 3 = (3/4)(x − 5), so 3x − 4y − 3 = 0 [1] (b) y = 0 gives C(1, 0) [1]. C is on the perpendicular bisector, so CM ⊥ AB, with CM = √(4² + 3²) = 5 and AB = 10 [1]. Area = ½ × 10 × 5 = 25 [1] Examiner insight: in (b) the height must be perpendicular to the base; using AC or BC as the height without justification loses the method mark.

4. (a) (x + 4)² − 16 + (y − 5)² − 25 + 16 = 0 [1], so (x + 4)² + (y − 5)² = 25. Centre (−4, 5) [1], radius 5 [1] (b) 1 + 1 − 8 − 10 + 16 = 0, so P is on the circle [1]. Gradient of radius CP = (1 − 5)/(−1 + 4) = −4/3 [1]. Tangent gradient = 3/4 [1]. y − 1 = (3/4)(x + 1), so 3x − 4y + 7 = 0 [1] Examiner insight: “show that” needs the substitution written out and a conclusion; a bare “it lies on the circle” scores zero.

5. (a) Centre = midpoint (3, 1) [1]. r² = 4² + 3² = 25 [1]. (x − 3)² + (y − 1)² = 25 [1] (b) (6 − 3)² + (5 − 1)² = 9 + 16 = 25, so C lies on the circle [1] (c) Gradient AC = 7/7 = 1 and gradient BC = 1/(−1) = −1; product = −1, so AC ⊥ BC [1]. The angle in a semicircle is a right angle [1] Examiner insight: both gradients and their product must be shown; stating “angle in a semicircle” without the gradient check earns only the second mark.

6. (a) y = 4 − x [1]. x² + (4 − x)² − 2x + 4(4 − x) − 20 = 0 gives 2x² − 14x + 12 = 0, so x² − 7x + 6 = 0 [1]. x = 1 or 6 [1]. P(1, 3), Q(6, −2) [1] (b) Midpoint of PQ is (7/2, 1/2); gradient PQ = −1, so the perpendicular gradient is 1 [1]. Perpendicular bisector: y = x − 3 [1]. The centre is (1, −2), and 1 − 3 = −2, so the centre lies on it [1] Examiner insight: the y-coordinates in (a) should come from the line; substituting x back into the circle gives extra values and can lose the final accuracy mark.

7. (a) x² − 5x + 7 = x + k gives x² − 6x + (7 − k) = 0 [1]. Tangent: 36 − 4(7 − k) = 0 [1], so k = −2 [1]. Then x² − 6x + 9 = 0, x = 3, y = 1: (3, 1) [1] (b) Discriminant 8 + 4k < 0: k < −2 [1] Examiner insight: the method mark needs discriminant = 0 applied to the combined quadratic, not to the curve alone.

8. x² + (2x + c)² = 20 gives 5x² + 4cx + c² − 20 = 0 [1]. Tangent: (4c)² − 4(5)(c² − 20) = 0 [1], so 400 − 4c² = 0 [1] and c = 10 or c = −10 [1]. c = 10: 5x² + 40x + 80 = 0, x = −4, point (−4, 2) [1]. c = −10: x = 4, point (4, −2) [1] Examiner insight: both values of c are needed for the accuracy mark; taking only the positive square root loses it and the point that follows.

9. (a) Midpoint (4, 3) [1]. Gradient AB = −2/6 = −1/3, so the perpendicular gradient is 3 [1]. y = 3x − 9 [1] (b) 3x − 9 = x − 3 [1], so centre (3, 0) [1]. r² = (1 − 3)² + (4 − 0)² = 20 [1]. (x − 3)² + y² = 20 [1] (c) Gradient of radius to A = 4/(−2) = −2 [1]. Tangent gradient = 1/2 [1]. y − 4 = (1/2)(x − 1), so x − 2y + 7 = 0 [1] (d) y = 0 gives T(−7, 0) [1]. TA = √(8² + 4²) = √80 = 4√5 [1] Examiner insight: errors in the centre carry forward; follow-through method marks are usually allowed in (c) and (d), but accuracy marks need correct values.

10. Let C be (c, 0). Gradient AC = −6/c, gradient BC = −2/(c − 8) [1]. Perpendicular: (−6/c)(−2/(c − 8)) = −1 [1], so 12 = −c(c − 8), giving c² − 8c + 12 = 0 [1]. (c − 2)(c − 6) = 0 [1]. C(2, 0) or C(6, 0) [1] Examiner insight: the circle with diameter AB, (x − 4)² + (y − 4)² = 20, meeting y = 0 is an equally valid method; either route earns full marks if complete.

11. (x − 5)² + m²x² = 9 [1] gives (1 + m²)x² − 10x + 16 = 0 [1]. Two points: 100 − 64(1 + m²) > 0 [1], so 36 − 64m² > 0, m² < 9/16 [1]. −3/4 < m < 3/4 [1] Examiner insight: the final mark needs the inequality in both directions; m < 3/4 alone, or m < ±3/4, is not accepted.

12. (a) The diagonals of a rhombus bisect each other at right angles, so BD passes through the midpoint (5, 4) of AC [1]. Gradient AC = 4/8 = 1/2, so gradient BD = −2 [1]. y = −2x + 14 [1] (b) x = 0 gives B(0, 14) [1]. The midpoint of BD is (5, 4), so D = (2 × 5 − 0, 2 × 4 − 14) [1] = (10, −6) [1] (c) AC = √80 = 4√5 and BD = √(10² + 20²) = √500 = 10√5 [1]. Area = ½ × AC × BD [1] = ½ × 4√5 × 10√5 = 100 [1] Examiner insight: the first mark in (a) needs the geometric reason stated; a line through the midpoint with no reason is harder to credit if the gradient then goes wrong.

Where marks are usually lost

• The perpendicular gradient given as −m instead of −1/m.
• A perpendicular bisector drawn through an end point instead of the midpoint.
• Answers not in the requested form ax + by + c = 0 with integer coefficients.
• Centre signs reversed when reading (x + a)² or completing the square.
• r² quoted as the radius, or a decimal given where an exact surd was asked for.
• The tangent built from the radius gradient itself, with no perpendicular step.
• The second coordinate of an intersection taken from the circle rather than the line.
• Only one value kept from ±√, losing a second tangent or a second point.
• “Show that” steps missing: substitution and conclusion must both be written.

Next steps

• Coordinate geometry revision notes
• Coordinate geometry study guide
• Pure Mathematics 1 mixed practice
• A Level Mathematics hub
• Printable 9709 checklist
• AS Level diagnostic, the 9709 self-check bank, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Section 1 Pure Mathematics 1 (for Paper 1): 1.3 Coordinate geometry.