OxfordAQA International AS and A-level Mathematics: Straight lines and intersections of lines with curves (9660) – 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 P1.2 (Coordinate geometry) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2, for International AS exams from May/June 2018 and A-level exams from May/June 2019. The content belongs to International AS Unit P1 (Pure Maths). If you get stuck, use the study guide and the revision notes. The course hub, the printable checklist and the free 10-minute diagnostics are also available. Give exact answers unless told otherwise.

Questions

1. The points A and B have coordinates (−4, 3) and (2, −1).

(a) Find the gradient of AB. [1] (b) Find the mid-point of AB. [1] (c) Find the length of AB, giving your answer as a surd in its simplest form. [2]

2. (a) Find the gradient and the y-intercept of the line 6x − 4y + 9 = 0. [2] (b) Find an equation of the line through (3, −2) with gradient −5/4, in the form ax + by + c = 0 where a, b and c are integers. [3]

3. The line l passes through P(−2, 9) and Q(6, −3).

(a) Find an equation of l in the form ax + by + c = 0, where a, b and c are integers. [3] (b) Find the coordinates of the points where l meets the axes. [2] (c) Find the area of the triangle bounded by l and the two axes. [2]

4. Three lines have equations L₁: 4x − 6y + 5 = 0, L₂: 3x + 2y − 8 = 0 and L₃: 2x − 3y = 1.

(a) Identify, with reasons, a pair of parallel lines and a pair of perpendicular lines. [4] (b) Find an equation of the line through (6, −2) perpendicular to L₁, in the form ax + by + c = 0. [3]

5. The points A and B have coordinates (1, −2) and (7, 6).

(a) Find an equation of the perpendicular bisector of AB, in the form ax + by + c = 0 with integer coefficients. [4] (b) The point C is (8, −1). Show that C lies on the perpendicular bisector, and show that CA = CB. [2]

6. The line y = 2x + 2 meets the curve y = x² − 3x − 4 at the points M and N.

(a) Find the coordinates of M and N. [4] (b) Find the mid-point of MN and the exact length of MN. [3]

7. Find the coordinates of the points where the curve y = 18/x meets the line 2x + y = 15. [5]

8. The line y = kx + 3 is a tangent to the curve y = x² + 4x + 7.

(a) Find the two possible values of k. [4] (b) Find the coordinates of the point of contact for each value of k. [3]

9. The line y = x + c and the curve y = 2x² − 3x + 1 are given. Find the values of c for which the line

(a) meets the curve at two distinct points, (b) is a tangent to the curve, (c) does not meet the curve. [5]

10. By solving a suitable equation, show that the line y = −2 touches the curve y = x³ − 3x at one point and crosses it at another, and give the coordinates of both points. [5]

11. The points P, Q and R have coordinates (1, 1), (5, 3) and (2, 9). The quadrilateral PQRS is a rectangle.

(a) Show that PQ is perpendicular to QR. [3] (b) Find the coordinates of S. [2] (c) Find an equation of the diagonal PR in the form ax + by + c = 0. [2] (d) Find the area of the rectangle. [2]

Answers

1. (a) Gradient = (−1 − 3)/(2 − (−4)) = −4/6 = −2/3 [1] (b) Mid-point = ((−4 + 2)/2, (3 − 1)/2) = (−1, 1) [1] (c) AB² = 6² + (−4)² = 36 + 16 = 52 [1], so AB = √52 = 2√13 [1] Examiner insight: Keep brackets round negative coordinates, and simplify the surd if the question asks for simplest form, because the question asks for a simplified surd, so leaving √52 would not answer it fully.

2. (a) Rearrange: 4y = 6x + 9, so y = (3/2)x + 9/4 [1]. Gradient 3/2, y-intercept 9/4 [1] (b) y + 2 = −(5/4)(x − 3) [1] 4y + 8 = −5x + 15 [1] 5x + 4y − 7 = 0 [1] Examiner insight: Showing the line in y − y₁ = m(x − x₁) form first makes your method visible and easier to check if the rearrangement into integers slips.

3. (a) Gradient = (−3 − 9)/(6 − (−2)) = −12/8 = −3/2 [1] y − 9 = −(3/2)(x + 2), so 2y − 18 = −3x − 6 [1] 3x + 2y − 12 = 0 [1] (b) When y = 0, x = 4. When x = 0, y = 6. [1] The points are (4, 0) and (0, 6) [1] (c) The triangle has base 4 and height 6 and a right angle at the origin [1], so area = ½ × 4 × 6 = 12 [1] Examiner insight: Check your equation by substituting the second point, and use the axis intercepts from your own equation so an early slip is easy to spot.

4. (a) L₁: y = (2/3)x + 5/6, gradient 2/3. L₂: y = −(3/2)x + 4, gradient −3/2. L₃: y = (2/3)x − 1/3, gradient 2/3. [2] L₁ and L₃ have equal gradients, so they are parallel [1]. L₁ and L₂ have gradients with product (2/3)(−3/2) = −1, so they are perpendicular (L₂ and L₃ are also perpendicular) [1] (b) Perpendicular gradient = −3/2 [1] y + 2 = −(3/2)(x − 6), so 2y + 4 = −3x + 18 [1] 3x + 2y − 14 = 0 [1] Examiner insight: Give the gradients as evidence for each conclusion. A statement that lines “look parallel” without gradients is unlikely to be enough.

5. (a) Mid-point = (4, 2) [1]. Gradient AB = 8/6 = 4/3 [1]. Perpendicular gradient = −3/4 [1] y − 2 = −(3/4)(x − 4), so 4y − 8 = −3x + 12, giving 3x + 4y − 20 = 0 [1] (b) 3(8) + 4(−1) − 20 = 24 − 4 − 20 = 0, so C lies on the line [1]. CA² = 7² + 1² = 50 and CB² = 1² + 7² = 50, so CA = CB [1] Examiner insight: In a “show that” part, write the substitution line and the conclusion explicitly rather than just stating that the answer is zero.

6. (a) x² − 3x − 4 = 2x + 2 [1] x² − 5x − 6 = 0, so (x − 6)(x + 1) = 0 [1] x = 6 or x = −1 [1] Using y = 2x + 2: M(−1, 0) and N(6, 14) [1] (b) Mid-point = (5/2, 7) [1]. MN² = 7² + 14² = 245 [1], so MN = 7√5 [1] Examiner insight: Substitute the x-values into the line, not the curve, and check them in the curve afterwards. A coordinate pair is needed for each point.

7. Substitute y = 15 − 2x: 18/x = 15 − 2x [1] 18 = 15x − 2x², so 2x² − 15x + 18 = 0 [1] (2x − 3)(x − 6) = 0 [1] x = 3/2 or x = 6 [1] y = 15 − 2x gives (3/2, 12) and (6, 3) [1] Examiner insight: Multiply through by x only after you have written one equation in x. Give both coordinates for both points.

8. (a) x² + 4x + 7 = kx + 3, so x² + (4 − k)x + 4 = 0 [1] Equal roots: (4 − k)² − 16 = 0 [1] (4 − k)² = 16, so 4 − k = ±4 [1] k = 0 or k = 8 [1] (b) The repeated root is x = −(4 − k)/2. For k = 0, x = −2 and y = 3, giving (−2, 3) [1]. For k = 8, x = 2 and y = 8(2) + 3 = 19, giving (2, 19) [1]. Check: both points satisfy y = x² + 4x + 7 [1] Examiner insight: The tangent condition comes from the discriminant of the combined quadratic. Writing “equal roots” before setting it to zero shows your reasoning.

9. x + c = 2x² − 3x + 1, so 2x² − 4x + (1 − c) = 0 [1] Discriminant = 16 − 8(1 − c) = 8 + 8c [1] (a) Two distinct points: 8 + 8c > 0, so c > −1 [1] (b) Tangent: 8 + 8c = 0, so c = −1 [1] (c) No intersection: 8 + 8c < 0, so c < −1 [1] Examiner insight: Mark the strict or equals sign for each case carefully. Tangent is equal to zero, not greater than.

10. x³ − 3x = −2, so x³ − 3x + 2 = 0 [1] x = 1 gives 1 − 3 + 2 = 0, so (x − 1) is a factor, and x³ − 3x + 2 = (x − 1)²(x + 2) [2] The repeated root x = 1 means the line touches the curve at (1, −2) [1]. The single root x = −2 means it crosses at (−2, −2) [1] Examiner insight: Say what the repeated root means in words, not only the roots. A line through a point where the curve turns is the usual picture for a touch.

11. (a) Gradient PQ = 2/4 = 1/2 [1]. Gradient QR = 6/(−3) = −2 [1]. Product = −1, so PQ is perpendicular to QR [1] (b) Mid-point of PR = (3/2, 5) [1]. Diagonals bisect each other, so the mid-point of QS is (3/2, 5), giving S = (3 − 5, 10 − 3) = (−2, 7) [1] (c) Gradient PR = 8/1 = 8 [1]. y − 1 = 8(x − 1), so 8x − y − 7 = 0 [1] (d) PQ = √(16 + 4) = 2√5 and QR = √(9 + 36) = 3√5 [1]. Area = 2√5 × 3√5 = 30 [1] Examiner insight: Use the earlier parts. Finding S from the diagonal mid-point is quicker than solving gradient conditions, and it reuses the mid-point formula.

Where marks are usually lost

• Mixing the order of subtraction in the gradient formula so the sign is wrong.
• Reading the gradient of ax + by + c = 0 as a/b instead of −a/b.
• Using the reciprocal of the gradient without changing its sign for a perpendicular line.
• Using a point other than the mid-point when building a perpendicular bisector.
• Giving only x-values for points of intersection.
• Substituting into the curve instead of the line, which makes the arithmetic harder.
• Applying the discriminant to the curve itself instead of the quadratic after substitution.
• Writing “greater than 0” for a tangent or “equal to 0” for two intersections.
• Expanding (4 − k)² incorrectly, or dividing only some terms when clearing fractions.
• Leaving answers as decimals when exact surds or fractions were expected.

Next steps

• Revisit the revision notes for formulas and method boxes.
• Reread the study guide for worked examples.
• Go to the OxfordAQA Mathematics hub or tick topics off in the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class for help with any question you could not finish.

Official syllabus

These questions follow section P1.2 (Coordinate geometry) in section 3.1 of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2, published by OxfordAQA.