OCR GCSE Mathematics: Straight-line graphs, parallel and perpendicular lines, and equations of circles (J560)

This guide follows the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). It covers straight-line graphs (7.02a), parallel and perpendicular lines (7.02b) and the equation of a circle with centre at the origin (7.01f). Gradient, intercept, sketching, equations of lines and parallel lines are Foundation and Higher content. Inequality regions, perpendicular lines, tangents and circles are marked Higher tier only.

Use the revision notes for quick recall and the practice questions to test yourself. The OCR GCSE Mathematics hub, the checklist, the exam preparation guide and the 10-minute diagnostics help you plan.

What this guide covers

Ref.	What you must be able to do	Tier
7.02a	Find and interpret gradient and intercept, from a graph and from y = mx + c	Both
7.02a	Use y = mx + c to find and sketch lines	Both
7.02a	Write down the equation of a line when you are given two of its points, or one point and the gradient	Both
7.02b	Identify and find equations of parallel lines	Both
7.02a	Identify solution sets of linear inequalities in two variables, using dashed and solid lines	Higher tier only
7.02b	Identify and find equations of perpendicular lines	Higher tier only
7.02b	Work out the equation of the tangent where a circle is touched at a stated point	Higher tier only
7.01f	Recognise and use the equation of a circle with centre at the origin	Higher tier only

Gradient and intercept (7.02a)

The gradient m measures steepness. Divide the vertical change by the horizontal change between any two points: m = (change in y) ÷ (change in x). Lines that climb as you move right have positive m. Lines that fall have negative m. The y-intercept c is where the line crosses the y-axis, at the point (0, c).

A line written as y = mx + c shows both at once. Read m as the number multiplying x and c as the constant. If the equation is not in that form, rearrange it first.

Worked example 1. A line passes through (0, −2) and (3, 4). Find its equation.

Gradient m = (4 − (−2)) ÷ (3 − 0) = 6 ÷ 3 = 2
y-intercept c = −2 (the line crosses the y-axis at (0, −2))
Equation: y = 2x − 2

Worked example 2. Find the gradient and y-intercept of 3x + 4y = 20.

4y = −3x + 20
y = −(3/4)x + 5
Gradient = −3/4, y-intercept = 5

Interpreting a line. In a real context, the gradient tells you how fast one quantity changes for each unit of the other, and the intercept tells you the value at the start. A candle burns so that its height h cm after t hours is h = 24 − 1.5t. The gradient −1.5 means the height falls by 1.5 cm every hour. The intercept 24 is the height at the start. The candle is used up when 24 − 1.5t = 0, which gives t = 16 hours.

Sketching lines (7.02a)

A sketch needs the correct slope direction and the key points marked. You do not need to plot a table of values.

Method A: from y = mx + c. Mark (0, c). From there, move 1 across and m up (or down, if m is negative). Draw the line through both points and extend it.

Method B: find both intercepts. Put x = 0 to find where the line crosses the y-axis. Put y = 0 to find where it crosses the x-axis. This is quicker when the equation is in the form ax + by = d.

Worked example 3. Sketch y = −2x + 6.

y-intercept: (0, 6)
Gradient −2: across 1, down 2, so (1, 4), then (2, 2), then (3, 0)

The line falls steeply from (0, 6) and crosses the x-axis at (3, 0).

Worked example 4. Sketch 3x − 2y = 12.

x = 0:  −2y = 12, so y = −6    point (0, −6)
y = 0:  3x = 12, so x = 4      point (4, 0)

Join (0, −6) and (4, 0). The line rises to the right, so the gradient is positive. Check: 3x − 2y = 12 gives y = (3/2)x − 6, with gradient 3/2.

Two special cases. The line y = 5 is horizontal, with gradient 0. The line x = −2 is vertical, so it has no gradient value and cannot be written as y = mx + c.

Equation of a line (7.02a)

From two points. Find the gradient, then substitute one point to find c.

Worked example 5. Find the equation of the line through (1, −1) and (5, 5).

m = (5 − (−1)) ÷ (5 − 1) = 6 ÷ 4 = 3/2
Substitute (1, −1) into y = mx + c:  −1 = 3/2 + c, so c = −5/2
Equation: y = (3/2)x − 5/2

A different form is y − y₁ = m(x − x₁). It gives y + 1 = (3/2)(x − 1), which simplifies to the same line. Check with the other point: (3/2) × 5 − 5/2 = 5. Correct.

From a point and a gradient. Use the same idea with m already known.

Worked example 6. A line has gradient −4 and passes through (−3, 2).

y − 2 = −4(x − (−3))
y − 2 = −4x − 12
y = −4x − 10

Be careful with the signs when x₁ is negative: x − (−3) is x + 3.

Parallel lines (7.02b)

Parallel lines have equal gradients. Lines with different gradients meet at a point. To compare gradients, rearrange each equation to y = mx + c.

Worked example 7. Which of these lines are parallel?

A: y = (1/2)x + 3
B: 2y = x − 8        →  y = (1/2)x − 4
C: x − 2y = 4        →  y = (1/2)x − 2
D: 2x − y = 1        →  y = 2x − 1

A, B and C all have gradient 1/2, so they are parallel. D has gradient 2.

Worked example 8. Find the equation of the line through (2, 5) that is parallel to y = 3x − 7.

Parallel, so m = 3
5 = 3 × 2 + c, so c = −1
Equation: y = 3x − 1

Inequality regions (7.02a, Higher tier only)

A linear inequality such as y > x − 2 describes every point on one side of a line. The boundary line is y = x − 2. The convention is:

Symbol	Boundary line	Points on the line
< or >	Dashed	Not in the region
≤ or ≥	Solid	In the region

To decide which side, substitute a test point that is not on the line. The origin is easiest, unless the line passes through it. Check what the question asks you to shade: some questions shade the wanted region, others shade the unwanted region and leave the answer unshaded.

Worked example 9. The region R satisfies y > x − 2, x + y ≤ 6 and x ≥ 0. Describe R and list the points with integer coordinates on the line x = 2.

y > x − 2   dashed line through (0, −2) and (2, 0). Test (0, 0): 0 > −2 is true, so the side containing the origin.
x + y ≤ 6   solid line through (0, 6) and (6, 0). Test (0, 0): 0 ≤ 6 is true, so the side containing the origin.
x ≥ 0       solid line along the y-axis, with the region to its right.

R is a triangle with corners (0, −2), (0, 6) and (4, 2). The corner (4, 2) comes from solving y = x − 2 and x + y = 6. The sloping edge y = x − 2 is dashed, so points on it are not in R. The other two edges are solid.

On x = 2 you need y > 0 and y ≤ 4. The points are (2, 1), (2, 2), (2, 3) and (2, 4).

Writing inequalities from a region. Find the equation of each boundary line. Pick a point inside the region and test it in each inequality. Choose the symbol that makes the test true, using a strict symbol for a dashed line.

Perpendicular lines (7.02b, Higher tier only)

Two lines are perpendicular when their gradients multiply to give −1. If one gradient is m, the other is −1/m. This is the negative reciprocal: flip the fraction and change the sign.

Gradient	Perpendicular gradient
2	−1/2
−3/4	4/3
5	−1/5

Worked example 10. Find the equation of the line through (2, −3) that is perpendicular to y = (2/5)x + 1.

Gradient of the given line = 2/5
Perpendicular gradient = −5/2
−3 = (−5/2) × 2 + c, so c = −3 + 5 = 2
Equation: y = −(5/2)x + 2

Worked example 11. Find the gradient of a line perpendicular to 3x − 4y = 8.

−4y = −3x + 8, so y = (3/4)x − 2
Gradient 3/4, so the perpendicular gradient is −4/3

The equation of a circle (7.01f, Higher tier only)

A circle with centre at the origin (0, 0) and radius r has the equation

x² + y² = r²

This comes from Pythagoras. Every point (x, y) on the circle lies at distance r from the origin, so the right-angled triangle with sides x, y and r gives x² + y² = r². The number on the right is the radius squared.

• To find r from the equation, take the square root. The circle x² + y² = 40 has r = √40 = 2√10.
• To find the equation from a point on the circle, put the point’s coordinates in x² + y² to get r². A circle through (−5, 12) has r² = 25 + 144 = 169, so x² + y² = 169 and r = 13.
• To test a point, work out x² + y². The point is on the circle if the total equals r². It is inside if the total is less, and outside if it is more. For x² + y² = 40, the point (6, 2) gives 36 + 4 = 40, so it lies on the circle.

Tangent to a circle (7.02b, Higher tier only)

A tangent touches the circle at one point. It is perpendicular to the radius drawn to that point. So the method combines gradient, perpendicular gradient and a line equation.

Worked example 12. Find the equation of the tangent to x² + y² = 26 at the point (5, −1).

Check: 5² + (−1)² = 25 + 1 = 26, so the point is on the circle.
Gradient of the radius from (0, 0) to (5, −1) = −1/5
Tangent gradient = 5 (negative reciprocal)
y − (−1) = 5(x − 5)
y + 1 = 5x − 25
y = 5x − 26

Common errors

• Reading the gradient from the run over the rise. It is the rise over the run.
• Dropping a sign when rearranging, for example turning 3x + 4y = 20 into y = (3/4)x + 5 instead of y = −(3/4)x + 5.
• Treating 2y = x − 8 as having gradient 1. Divide every term by 2 first.
• Using a solid line for < or > in an inequality, or a dashed line for ≤ or ≥.
• Using the reciprocal without changing the sign, or changing the sign without flipping the fraction.
• Writing x² + y² = r for a circle. The right-hand side is r².
• Using the circle equation to get the tangent gradient. The tangent gradient comes from the radius.

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.