OCR GCSE Mathematics: Straight-line graphs, parallel and perpendicular lines, and equations of circles (J560) – 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 follow sections 7.01f, 7.02a and 7.02b of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Questions 1 to 6 are Foundation and Higher content. Questions 7 to 12 are (Higher tier only). Learn the methods first from the study guide and the revision notes. The OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics help with planning.

Questions

1. Find the gradient and the y-intercept of each line.

(a) y = 5 − 2x [2] (b) 4x + 2y = 7 [3]

2. A tank is filling with water. The graph of volume V litres against time t minutes is a straight line through (0, 12) and (5, 42).

(a) Find the equation of the line. [3] (b) Explain what the gradient and the y-intercept mean for the tank. [2]

3. Sketch the line 5x + 2y = 10. Mark the points where it crosses each axis. [4]

4. A line passes through (−2, 9) and (4, −3).

(a) Find its equation. [4] (b) Does the point (10, −15) lie on the line? Show your working. [1]

5. Find the equation of the line with gradient 2/3 that passes through (6, −1). [3]

6. Three lines have these equations.

A: y = 4 − 3x
B: 3x + y = 8
C: 6x − 2y = 1

(a) Which two lines are parallel? Give a reason. [2] (b) A line parallel to 2y = 6x + 1 passes through (−3, 4). Find its equation. [3]

7. (Higher tier only) The region R satisfies y > 1, x + 2y ≤ 10 and y ≤ 2x.

(a) State whether each boundary line is dashed or solid. [3] (b) Find the number of points with integer coordinates inside R, including any on a solid boundary. [4]

8. (Higher tier only) A triangle has corners (1, 1), (5, 1) and (5, 5). Its edge from (1, 1) to (5, 1) is dashed. The other two edges are solid. The region R is the inside of the triangle, with points on the solid edges but not on the dashed edge. Write three inequalities that define R. [3]

9. (Higher tier only) Line L has equation 2x + 5y = 10. A second line goes through (4, −3) at right angles to L. Find its equation. [4]

10. (Higher tier only) The points are A(−1, 2), B(3, 4) and C(4, 2).

(a) Show that angle ABC is a right angle. [3] (b) Find the equation of the line through C that is parallel to AB. [2]

11. (Higher tier only) A circle has equation x² + y² = 65.

(a) Show that the point (4, 7) lies on the circle. [2] (b) Write down the exact radius of the circle. [1] (c) Is the point (6, 5) inside, on or outside the circle? Show your working. [2] (d) Find the y-coordinates of the points on the circle where x = 1. [2]

12. (Higher tier only) The point P(3, −5) is on the circle x² + y² = 34.

(a) Show that P lies on the circle. [1] (b) Find the gradient of the radius from the origin to P. [1] (c) Find the equation of the tangent to the circle at P. [4] (d) The tangent crosses the x-axis at T. Find the x-coordinate of T. [1]

Answers

1. (a) Write as y = −2x + 5. [1] Gradient −2, y-intercept 5. [1] (b) 2y = −4x + 7 [1], so y = −2x + 7/2. [1] Gradient −2, y-intercept 7/2 (3.5). [1] Examiner insight: Answers are easier to earn when you rearrange visibly into y = mx + c. A common loss is to read the gradient as 4 straight from 4x + 2y = 7 without dividing by 2.

2. (a) Gradient = (42 − 12) ÷ (5 − 0) [1] = 6. [1] The y-intercept is 12, so V = 6t + 12. [1] (b) The gradient: 6 litres are added every minute. [1] The intercept: there were 12 litres in the tank at the start. [1] Examiner insight: An interpretation needs the units and the context. “The gradient is 6” alone does not say what changes by 6 each minute.

3. When x = 0: 2y = 10, so y = 5, giving the point (0, 5). [1] When y = 0: 5x = 10, so x = 2, giving the point (2, 0). [1] Draw a straight line through (0, 5) and (2, 0), extended past both axes. [1] The line slopes down to the right, which matches the gradient of −5/2 from y = −(5/2)x + 5. [1] Examiner insight: A sketch should show the line crossing the axes at labelled values. A line with no marked intercepts leaves the examiner nothing to check, and a line sloping the wrong way is a clear error.

4. (a) m = (−3 − 9) ÷ (4 − (−2)) [1] = −12 ÷ 6 = −2. [1] Substitute (−2, 9): 9 = −2 × (−2) + c, so c = 5. [1] y = −2x + 5 [1] (b) When x = 10: −2 × 10 + 5 = −15, which equals the y-coordinate, so yes, (10, −15) lies on the line. [1] Examiner insight: Subtract the coordinates in the same order on the top and the bottom. Checking the second given point in your equation is a quick way to catch a sign slip.

5. y − (−1) = (2/3)(x − 6) [1], so y + 1 = (2/3)x − 4. [1] y = (2/3)x − 5 [1] Examiner insight: Keep brackets around a negative y-value when you substitute. Many errors in this kind of question start with y − (−1) written as y − 1.

6. (a) A: gradient −3. B: y = 8 − 3x, gradient −3. C: y = 3x − 1/2, gradient 3. [1] A and B are parallel because they have the same gradient, −3. [1] (b) 2y = 6x + 1 gives y = 3x + 1/2, so m = 3. [1] Then 4 = 3 × (−3) + c, so c = 13. [1] y = 3x + 13 [1] Examiner insight: Lines B and C are a trap, because the numbers 3 and 6 look similar. State the gradients you found, since a reason with no gradients named is weak.

7. (a) y > 1 is dashed. [1] x + 2y ≤ 10 is solid. [1] y ≤ 2x is solid. [1] (b) Corners of R: (1/2, 1), (8, 1) and (2, 4). Take each whole-number y in turn. [1] For y = 2: x ≥ 1 and x ≤ 6, so 6 points. For y = 3: x ≥ 1.5 and x ≤ 4, so x = 2, 3, 4, which is 3 points. [1] For y = 4: x ≥ 2 and x ≤ 2, so 1 point (2, 4). [1] For y = 1 none are allowed, since y > 1. Total 6 + 3 + 1 = 10 points. [1] Examiner insight: Work along each row of the grid and write down what you find. Counting points from a rough picture is where most slips occur, especially on the boundaries.

8. The bottom edge is y = 1 and is dashed, so y > 1. [1] The right edge is x = 5 and is solid, so x ≤ 5. [1] The diagonal edge is y = x and is solid. Test (4, 2) inside the triangle: 2 ≤ 4, so y ≤ x. [1] Examiner insight: Test a point that is clearly inside the region against each inequality. Choosing the sign by guessing which side looks right often reverses it.

9. 5y = −2x + 10, so y = −(2/5)x + 2 and the gradient of L is −2/5. [1] The perpendicular gradient is 5/2. [1] Substitute (4, −3): −3 = (5/2) × 4 + c, so c = −13. [1] y = (5/2)x − 13 [1] Examiner insight: Rearrange L to y = mx + c before taking the negative reciprocal. Taking the reciprocal of the coefficient of x in 2x + 5y = 10 gives the wrong gradient.

10. (a) Gradient of AB = (4 − 2) ÷ (3 − (−1)) = 2/4 = 1/2. [1] Gradient of BC = (2 − 4) ÷ (4 − 3) = −2. [1] (1/2) × (−2) = −1, so AB and BC are perpendicular and angle ABC = 90°. [1] (b) The gradient is 1/2. [1] Substitute (4, 2): 2 = 2 + c, so c = 0. y = x/2 [1] Examiner insight: For a “show that” question, work out both gradients, write their product and state the conclusion. Saying the lines “look perpendicular” does not prove anything.

11. (a) 4² + 7² = 16 + 49 [1] = 65, so the point lies on the circle. [1] (b) r = √65 [1] (c) 6² + 5² = 36 + 25 = 61, which is less than 65, so (6, 5) is inside the circle. [1] [1] (d) 1 + y² = 65, so y² = 64. [1] y = 8 or −8 [1] Examiner insight: Give both y-values in part (d). Taking only the positive root loses the second point. In part (a), show the sum and the total, not only the word “yes”.

12. (a) 3² + (−5)² = 9 + 25 = 34, so P is on the circle. [1] (b) Gradient of the radius = −5/3. [1] (c) The tangent is perpendicular to the radius, so its gradient is 3/5. [1] y − (−5) = (3/5)(x − 3) [1], so y + 5 = (3/5)x − 9/5. [1] y = (3/5)x − 34/5 (or 3x − 5y = 34) [1] (d) At T, y = 0: 0 = (3/5)x − 34/5, so x = 34/3. [1] Examiner insight: Tangent questions reward a clear chain: radius gradient, negative reciprocal, then the line through the point. An exact fraction such as 34/3 is safer than a rounded decimal.

Where marks are usually lost

• Dividing only some of the terms when rearranging to y = mx + c, so the gradient is wrong.
• Mixing up the order of subtraction in the gradient, with one difference reversed.
• Losing a negative sign when substituting a point with a negative coordinate.
• Giving interpretation answers with no units or no context.
• Sketching a line with no marked intercepts or the wrong direction of slope.
• Choosing solid or dashed lines from habit rather than from the sign in the inequality.
• Taking the reciprocal but not changing the sign for a perpendicular gradient, or the reverse.
• Writing x² + y² = r instead of r², or giving only the positive root when a circle is cut at two points.
• Using the gradient of the radius as the gradient of the tangent.

Next steps

• Re-read the revision notes for any method that cost you marks.
• Go back to the study guide for the worked examples.
• Plan your work with the OCR GCSE Mathematics hub and the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Sections 7.01f, 7.02a and 7.02b.