AQA A-Level Mathematics: C: Coordinate geometry in the (x, y) plane (7357) – 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 C: Coordinate geometry in the (x, y) plane of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards: content references C1 to C4. Section C is listed under Paper 1, and Papers 2 and 3 can also assess it. A calculator is required in every 7357 assessment, but show your algebra: the method marks are for the working. Questions 11 and 12 are longer problems that combine ideas.

Learn the methods first in the study guide and the revision notes. Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/.

Questions

1. Find the equation of the line through (3, −4) that is parallel to 5x − 2y + 7 = 0. Give your answer in the form ax + by + c = 0, where a, b and c are integers. [3]

2. The lines L₁ and L₂ have equations 2x − 3y + 13 = 0 and 3x + 2y − 13 = 0.

(a) Show that L₁ and L₂ are perpendicular. [2] (b) Find the coordinates of P, the point where L₁ and L₂ meet. [3] (c) Find the exact area of the triangle formed by L₁, L₂ and the y-axis. [3]

3. A plumber charges a fixed call-out fee plus an hourly rate. A 2-hour job costs £130 and a 5-hour job costs £265. Assume the cost £C is a linear function of the time h hours.

(a) Find C in terms of h. [3] (b) Interpret the gradient and the intercept of your model in context. [2]

4. A circle has equation x² + y² + 8x − 2y − 8 = 0.

(a) Find the centre and radius of the circle. [3] (b) Determine whether the point (−1, 4) lies inside, on or outside the circle. [2]

5. A circle C has equation x² + y² − 10x − 4y + 4 = 0. The points A(1, −1) and B(8, 6) lie on C.

(a) Find the centre and radius of C. [3] (b) M is the midpoint of AB. Show that the line from the centre to M is perpendicular to AB. [3] (c) Find the exact shortest distance from the centre of C to the chord AB. [2]

6. P(−3, 2) and Q(5, 8) are the ends of a diameter of a circle.

(a) Find the equation of the circle. [3] (b) Verify that R(4, 9) lies on the circle. [1] (c) State the size of angle PRQ, giving a reason, and hence find the area of triangle PQR. [3]

7. The line y = kx + 5 is a tangent to the circle x² + y² = 5.

(a) Find the possible values of k. [4] (b) For the positive value of k, find the point of contact. [2]

8.

(a) A curve has parametric equations x = 4cos θ, y = 3sin θ, 0 ≤ θ < 2π. Find its Cartesian equation. [3] (b) Write down parametric equations for the circle (x + 1)² + (y − 4)² = 49. [2]

9. A curve has parametric equations x = 2 + sin θ, y = cos²θ, 0 ≤ θ < 2π. Find its Cartesian equation and state the set of possible values of x. [4]

10. A curve has parametric equations x = t + 2, y = t² − 1.

(a) Find a Cartesian equation of the curve. [2] (b) Find the coordinates of the points where the curve meets the line y = 4x − 12. [4]

11. A point P on the rim of an observation wheel is modelled by x = 18sin(πt/12), y = 20 − 18cos(πt/12), where t is the time in minutes from when P is at its lowest point, and x and y are in metres, with y the height above the ground.

(a) Find a Cartesian equation of the path of P. [3] (b) State the radius of the wheel and the height of its centre. [1] (c) Find the time for one revolution. [1] (d) Find the first time at which P is 29 m above the ground. [3] (e) Find how long P spends above 29 m in each revolution. [1] (f) Give one limitation of the model. [1]

12. A circle has centre C(1, 3) and equation (x − 1)² + (y − 3)² = 20.

(a) Show that A(5, 5) and B(3, −1) lie on the circle. [2] (b) Find the equations of the tangents to the circle at A and at B. [4] (c) The tangents meet at T. Find the coordinates of T. [2] (d) Show that CATB is a square and find its area. [3]

Answers

1. Gradient of 5x − 2y + 7 = 0 is 5/2 [1]. y + 4 = (5/2)(x − 3) [1], so 2y + 8 = 5x − 15, giving 5x − 2y − 23 = 0 [1]. Examiner insight: the final mark needs the exact form asked for; y = (5/2)x − 23/2 is correct but does not have integer a, b, c in ax + by + c = 0, so it loses the last mark.

2. (a) Gradients are 2/3 and −3/2 [1]. (2/3)(−3/2) = −1, so perpendicular [1]. (b) Eliminate y: 4x − 6y = −26 and 9x + 6y = 39 [1], so 13x = 13 [1]. P = (1, 5) [1]. (c) y-intercepts: 13/3 (L₁) and 13/2 (L₂) [1]. Base = 13/2 − 13/3 = 13/6, height = 1 (the x-coordinate of P) [1]. Area = (1/2)(13/6)(1) = 13/12 [1]. Examiner insight: in a “show that”, state the conclusion (“so the lines are perpendicular”) after the product; a bare −1 with no statement can lose the final mark.

3. (a) Gradient = (265 − 130)/(5 − 2) = 45 [1]. Intercept = 130 − 2 × 45 = 40 [1]. C = 45h + 40 [1]. (b) £45 is the charge for each hour [1]; £40 is the fixed call-out fee [1]. Examiner insight: interpretations must use the context and units (“£45 per hour”), not “the gradient is the rate of change”.

4. (a) (x + 4)² − 16 + (y − 1)² − 1 − 8 = 0 [1]. (x + 4)² + (y − 1)² = 25 [1]. Centre (−4, 1), radius 5 [1]. (b) (−1 + 4)² + (4 − 1)² = 9 + 9 = 18 [1]. 18 < 25, so the point is inside [1]. Examiner insight: the comparison must be with r² = 25 (or √18 with r = 5); comparing 18 with 5 is a method error and scores nothing for the conclusion.

5. (a) (x − 5)² − 25 + (y − 2)² − 4 + 4 = 0 [1]. Centre (5, 2) [1], radius 5 [1]. (b) M = (9/2, 5/2) [1]. Gradient AB = 7/7 = 1 and gradient from (5, 2) to M = (1/2)/(−1/2) = −1 [1]. Product −1, so the line is perpendicular to AB [1]. (c) The perpendicular from the centre meets AB at M, so the distance is √((1/2)² + (1/2)²) [1] = √2/2 [1]. Examiner insight: “exact” means a surd; 0.707 earns the method mark but not the accuracy mark.

6. (a) Centre = midpoint of PQ = (1, 5) [1]. PQ = √(8² + 6²) = 10, so r = 5 [1]. (x − 1)² + (y − 5)² = 25 [1]. (b) (4 − 1)² + (9 − 5)² = 9 + 16 = 25, so R lies on the circle [1]. (c) 90°, because the angle in a semicircle is a right angle [1]. PR = √(49 + 49) = 7√2 and QR = √(1 + 1) = √2 [1]. Area = (1/2)(7√2)(√2) = 7 [1]. Examiner insight: the reason mark needs the circle property linked to PQ being a diameter; a bare “90°” with no reason does not earn it.

7. (a) x² + (kx + 5)² = 5 [1]. (1 + k²)x² + 10kx + 20 = 0 [1]. Tangent, so 100k² − 80(1 + k²) = 0 [1], giving 20k² = 80 and k = ±2 [1]. (b) k = 2: 5x² + 20x + 20 = 0, so x = −2 [1]. y = 2(−2) + 5 = 1: (−2, 1) [1]. Examiner insight: the method mark for the discriminant needs “= 0” for a tangent; using “> 0” or “≥ 0” loses that mark and everything that follows from it.

8. (a) cos θ = x/4 and sin θ = y/3 [1]. cos²θ + sin²θ = 1 [1], so x²/16 + y²/9 = 1 [1]. (b) x = −1 + 7cos θ [1], y = 4 + 7sin θ, 0 ≤ θ < 2π [1]. Examiner insight: writing 49 instead of 7 as the multiplier is a common slip; the coefficient is r, not r².

9. sin θ = x − 2 [1]. y = cos²θ = 1 − sin²θ [1], so y = 1 − (x − 2)² [1]. Since −1 ≤ sin θ ≤ 1, 1 ≤ x ≤ 3 [1]. Examiner insight: the domain is a separate mark; the Cartesian equation alone scores at most 3 out of 4.

10. (a) t = x − 2 [1], so y = (x − 2)² − 1 [1]. (b) t² − 1 = 4(t + 2) − 12 [1]. t² − 4t + 3 = 0 [1], so t = 1 or t = 3 [1]. Points (3, 0) and (5, 8) [1]. Examiner insight: the final mark needs both points as coordinates; stopping at the values of t is incomplete.

11. (a) sin(πt/12) = x/18 and cos(πt/12) = (20 − y)/18 [1]. Using sin² + cos² = 1 [1]: x² + (y − 20)² = 324 [1]. (b) Radius 18 m; centre 20 m above the ground [1]. (c) One revolution when πt/12 = 2π, so t = 24 minutes [1]. (d) 20 − 18cos(πt/12) = 29 [1]. cos(πt/12) = −1/2 [1]. πt/12 = 2π/3, so t = 8 minutes [1]. (e) The next solution is πt/12 = 4π/3, t = 16, so P is above 29 m for 8 minutes [1]. (f) For example, the wheel may stop or slow down to let people on and off, so the speed is not constant [1]. Examiner insight: in (d), angles in a model like this must be in radians; working in degrees with π in the equation gives a wrong time and loses the accuracy mark.

12. (a) A: 4² + 2² = 20 [1]. B: 2² + (−4)² = 20 [1]. (b) Gradient CA = 2/4 = 1/2, so tangent gradient −2 [1]: y = −2x + 15 [1]. Gradient CB = −4/2 = −2, so tangent gradient 1/2 [1]: x − 2y − 5 = 0 [1]. (c) x − 2(−2x + 15) − 5 = 0 [1], so x = 7 and T = (7, 1) [1]. (d) (−2)(1/2) = −1, so the tangents are perpendicular; each is perpendicular to its radius, so all four angles of CATB are 90° [1]. CA = CB = √20 and TA = √(2² + 4²) = √20, so adjacent sides are equal [1]. Area = (√20)² = 20 [1]. Examiner insight: a “show that” for a square needs both right angles and equal adjacent sides; showing only one of them leaves the shape unproved and loses a mark.

Where marks are usually lost

• Giving a line in the wrong form, or with fractional a, b, c when integers were asked for.
• Taking the gradient of 5x − 2y + 7 = 0 as 5 or −5/2 instead of 5/2.
• Sign errors in the centre: (x + 4)² means the centre has x = −4.
• Comparing a distance with r² (or a squared distance with r) when testing inside or outside.
• Using the radius gradient as the tangent gradient.
• Not stating the circle property that justifies a right angle or a bisected chord.
• Using the circle equation instead of the line to find y after solving a line-circle quadratic.
• Leaving out the range of x after eliminating a trigonometric parameter.
• Working in degrees when the model is in radians.
• Giving a decimal when the question says “exact”.

Next steps

• Recap with the revision notes.
• Relearn any method in the study guide.
• Practise the discriminant with quadratics practice and parametric gradients with differentiation practice.
• See all 7357 pages on the course hub and tick off the checklist.
• Try one of the all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018 (A-level exams June 2018 onwards), section 3.4 C: Coordinate geometry in the (x, y) plane, content references C1 to C4. Published by AQA.