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

These notes condense sections 7.01f, 7.02a and 7.02b of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Read the full study guide for explanations, then test yourself with the practice questions. Items marked Higher tier only are not needed on the Foundation tier. See also the OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics.

Key facts

Idea	Rule	Tier
Gradient	m = (change in y) ÷ (change in x)	Both
Straight line	y = mx + c, with gradient m and y-intercept (0, c)	Both
Line with gradient m passing through the point (x₁, y₁)	y − y₁ = m(x − x₁)	Both
Parallel lines	Equal gradients	Both
Perpendicular lines	m₁ × m₂ = −1, so the gradient is −1/m	Higher tier only
Inequality boundary	Dashed for < or >, solid for ≤ or ≥	Higher tier only
Circle, centre origin	x² + y² = r²	Higher tier only
Tangent	Meets the circle once, at right angles to the radius	Higher tier only

Gradient and intercept (7.02a)

• Positive gradient: slopes up to the right. Negative gradient: slopes down.
• Rearrange to y = mx + c before reading off m. For 2y = 6x + 1, divide everything by 2 to get y = 3x + 1/2.
• In context, the gradient is the rate of change and c is the starting value.
• Horizontal line: y = b, gradient 0. Vertical line: x = a, no gradient value.

Reminder. The line 5x + 2y = 20 becomes 2y = −5x + 20, so y = −(5/2)x + 10. The gradient is −5/2 and the y-intercept is 10.

Context reminder. A pool is filling so that V = 40 + 15t litres after t minutes. The gradient 15 means 15 litres are added each minute. The intercept 40 means there were 40 litres at the start. After 6 minutes: V = 40 + 90 = 130 litres.

Sketching (7.02a)

Method in steps.

1. If the line is y = mx + c, plot (0, c).
2. Use the gradient to find a second point: across 1, then up m.
3. If the line is ax + by = d, find both intercepts instead: set x = 0, then set y = 0.
4. Draw a straight line through the points and extend it.
5. Label the intercepts.

Finding an equation (7.02a)

Two points, in steps.

1. Work out m = (y₂ − y₁) ÷ (x₂ − x₁).
2. Put one of the points into y = mx + c and solve for c.
3. Check with the other point.

One point and a gradient. Substitute straight into y − y₁ = m(x − x₁) and tidy up.

Reminder. Through (2, 8) and (6, 0): m = (0 − 8) ÷ (6 − 2) = −2. Then 8 = −4 + c gives c = 12. So y = −2x + 12. Check (6, 0): −12 + 12 = 0.

Parallel lines (7.02b)

• Rearrange both equations to y = mx + c and compare m.
• A line parallel to y = 4x − 3 has the form y = 4x + c. Put in the given point to find c.

Inequality regions (7.02a, Higher tier only)

Method in steps.

1. Draw the boundary line (replace the inequality sign with =).
2. Dashed line if the sign is < or >. Solid line if it is ≤ or ≥.
3. Test a point that is not on the line, for example (0, 0).
4. A true test means the region lies on the same side as your test point. A false test means it lies on the other side.
5. For several inequalities, the answer is where all conditions hold together.
6. Check what the question wants you to shade: the region that satisfies the inequalities, or the unwanted part.

Reminder. For y ≤ 3x − 1, the point (0, 0) gives 0 ≤ −1, which is false. The region is on the other side of the line from the origin, and the line is solid.

Reminder with two inequalities. The region satisfies y < 4 and y ≥ x. Draw y = 4 dashed and y = x solid. Test (1, 2): 2 < 4 is true and 2 ≥ 1 is true, so (1, 2) is in the region. Test (3, 1): 1 ≥ 3 is false, so (3, 1) is not in the region. A point on y = x, such as (2, 2), is in the region. A point on y = 4, such as (0, 4), is not.

Perpendicular lines (7.02b, Higher tier only)

• Flip the fraction and change the sign: 3 becomes −1/3, and −2/7 becomes 7/2.
• Test for perpendicular: multiply the gradients. The product must be exactly −1.
• Parallel and perpendicular compared: parallel needs the same m; perpendicular needs the negative reciprocal of m.

Reminder. Perpendicular to 2x − 3y = 12 through (6, 1). First, y = (2/3)x − 4, so m = 2/3. The perpendicular gradient is −3/2. Then 1 = −9 + c gives c = 10. So y = −(3/2)x + 10.

Circles and tangents (7.01f and 7.02b, Higher tier only)

Task	Method
Radius from equation	r = √(number on the right)
Equation from a point (a, b) on the circle	r² = a² + b²
Is a point on, inside or outside?	Compare x² + y² with r²
Tangent at (a, b)	Radius gradient b/a, then tangent gradient −a/b, then line through (a, b)

Tangent method in steps.

1. Check that the point is on the circle.
2. Find the gradient of the radius from the origin to the point.
3. Take the negative reciprocal.
4. Use y − y₁ = m(x − x₁) with the point of contact.

Reminder. Circle x² + y² = 89, point of contact (5, 8). Check: 25 + 64 = 89. Radius gradient 8/5, tangent gradient −5/8. Then y − 8 = −(5/8)(x − 5), which gives y = −(5/8)x + 89/8. Equivalent form: 5x + 8y = 89.

Final checks before you move on

• Does your equation pass through both of the points you were given? Substitute them.
• Is the answer in the form the question asks for, such as y = mx + c?
• Did you simplify fractions in the gradient, for example 6/4 to 3/2?
• For a tangent, did you check first that the point is on the circle?
• For a region, did you test a point rather than guess the side?

Must-know distinctions

• Gradient and intercept. The gradient is how steep. The intercept is where the line meets the y-axis.
• Dashed and solid. Strict inequalities use dashed lines, because points on the line are not included.
• r and r². The circle equation holds r², not r.
• Tangent and radius. The tangent gradient is the negative reciprocal of the radius gradient, never the radius gradient itself.

Quick self-test

1. State the gradient and y-intercept of y = 7 − 3x.
2. Find the gradient of the line 4x + 2y = 9.
3. Find the equation of the line through (0, 5) and (2, 11).
4. Find the equation of the line with gradient −1/2 through (4, 3).
5. Write the equation of a line parallel to y = 5x + 2 through (1, 9).
6. (Higher tier only) Is the boundary of y ≥ 2x + 1 solid or dashed? Is (0, 0) in the region?
7. (Higher tier only) Write the gradient of a line perpendicular to y = −(4/7)x + 3.
8. (Higher tier only) Does (3, 3) lie on, inside or outside x² + y² = 20?
9. (Higher tier only) Write the equation of a circle with centre the origin through (−3, 6).
10. (Higher tier only) Find the equation of the tangent to x² + y² = 29 at (5, 2).

Answers

1. Gradient −3, y-intercept 7.
2. 2y = −4x + 9, so y = −2x + 9/2. Gradient −2.
3. m = (11 − 5) ÷ 2 = 3 and c = 5, so y = 3x + 5.
4. y − 3 = −(1/2)(x − 4), so y = −(1/2)x + 5.
5. m = 5, and 9 = 5 + c gives c = 4. So y = 5x + 4.
6. Solid, because of ≥. At (0, 0): 0 ≥ 1 is false, so (0, 0) is not in the region.
7. 7/4.
8. 9 + 9 = 18, which is less than 20, so (3, 3) is inside.
9. r² = 9 + 36 = 45, so x² + y² = 45.
10. Check 25 + 4 = 29. Radius gradient 2/5, tangent gradient −5/2. Then y − 2 = −(5/2)(x − 5), so y = −(5/2)x + 29/2.

Where marks are usually lost

• Working out rise over run the wrong way round, or subtracting the coordinates in a different order on top and bottom.
• Giving the gradient of 2y = 6x + 1 as 6 because the equation was not divided through.
• Leaving the answer as y − y₁ = m(x − x₁) when the question asks for y = mx + c.
• Sketching a line without labelling any intercept or showing no scale on the points.
• Choosing a solid line for a strict inequality, or the wrong side after testing a point.
• Flipping the gradient but forgetting to change its sign for a perpendicular line.
• Writing a tangent gradient equal to the radius gradient.
• Giving a circle equation with r on the right rather than r².

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.