Line Equations — AQA GCSE Mathematics

SPEC A9, A10 • ALGEBRA — LINEAR GRAPHS

y = mx + c: Gradient, Intercept, Parallel & Perpendicular Lines

Two numbers — gradient m and y-intercept c — and you know everything about a straight line.

The Master Form: y = mx + c

Every non-vertical straight line has an equation of the form:

y = mx + c

m — Gradient

The gradient (steepness). Positive m → line slopes up. Negative m → slopes down. m = 0 → horizontal.

c — y-Intercept

Where the line crosses the y-axis. Set x = 0 in y = mx + c → y = c. So c is read off directly.

Why this matters

Once you know m and c, you can sketch the line, find any point on it, or compare it with another line — all without plotting a single value from a table.

Reading m and c From the Equation

If the equation is already in y = mx + c form, the gradient and intercept sit right there in the letters. Just match.

EquationGradient (m)y-Intercept (c)

y = 3x + 232

y = −4x + 1−41

y = x − 51−5

y = −x−10

y = 707 (horizontal line)

⚠️ Rearrange first. If the equation is not in y = … form — e.g. 2y = 6x + 4 or 3x + y = 5 — rearrange to isolate y before reading off m and c.

Example: 2y = 6x + 4 → y = 3x + 2, so m = 3, c = 2.

Example: 3x + y = 5 → y = −3x + 5, so m = −3, c = 5.

Gradient Graphically: Rise Over Run

On a graph, the gradient is rise ÷ run — how many units up for each unit across. Draw a "gradient triangle" to measure it.

Read Off the Gradient

Rise = 4, run = 2, so gradient m = 4 ÷ 2 = 2. Match this against y = 2x + 1 — the coefficient of x is 2 ✓. Either method gives the same answer.

Gradient From Two Points

When the line is given by two points rather than an equation, use the gradient formula:

Gradient Formula

m = (y₂ − y₁) / (x₂ − x₁)Change in y, divided by change in x.

Keep the order consistent

If you subtract y-coordinates as y₂ − y₁, you must subtract x-coordinates in the same order: x₂ − x₁. Swap either one and the sign of the gradient flips — and you get the wrong answer.

⚠️ Common arithmetic slip: students compute a difference like 22 − 16 = 8 (correct: 6). Check every subtraction — especially with negative coordinates. Source: AQA Jun23 P1 Q24 examiner report.

Worked Example — Gradient From Two Points01

A = (1, 2) and B = (4, 8). Find the gradient of AB. [2 marks]

M

Method: apply m = (y₂ − y₁)/(x₂ − x₁). Take A as (x₁, y₁) = (1, 2) and B as (x₂, y₂) = (4, 8).

C

Calculate: m = (8 − 2) / (4 − 1) = 6 / 3 = 2.

E

Evaluate: gradient = 2. Interpretation — for every 1 unit of x, y changes by 2. Line slopes up (positive gradient). M1 correct differences, A1 correct gradient.

Parallel Lines: Same Gradient

Parallel Rule

Parallel lines have the same gradient (same m) — but different y-intercepts (different c).

Example

The line y = 2x + 1 is parallel to y = 2x − 3 — both have gradient 2. They never meet because they slope at the same angle, just at different heights.

How to identify parallel lines

Given two equations, rearrange both into y = mx + c form first, then compare the gradients.

Example: Are y = 3x + 1 and 6x − 2y = 4 parallel?

Rearrange the second: 6x − 2y = 4 → −2y = −6x + 4 → y = 3x − 2. Gradient 3 ✓. Same gradient as y = 3x + 1, so yes — parallel.

Perpendicular Lines HIGHER

Perpendicular Rule

m₁ × m₂ = −1If two lines are perpendicular (meet at 90°), their gradients multiply to give −1.

Equivalently, m₂ = −1/m₁ — the negative reciprocal.

Two-step rule: "Flip AND negate"

Gradient 2 → flip to ½ → negate to −½. Check: 2 × (−½) = −1 ✓.

Gradient −⅔ → flip to −3/2 → negate to 3/2. Check: −⅔ × 3/2 = −1 ✓.

Point-Slope Form

Given a gradient m and one point (x₁, y₁) on the line, write the equation using:

y − y₁ = m(x − x₁)

Then expand and rearrange into y = mx + c form if required.

When to reach for this

Gradient + one point → use directly.

Two points → compute gradient first, then use this formula with either point.

Parallel line through a given point → copy the gradient, then use this formula.

Perpendicular line through a given pointHIGHER → find the negative reciprocal, then use this formula.

Why bother with point-slope?

It builds the equation in one mechanical step — no guessing or back-substituting for c. Once expanded and rearranged, you get y = mx + c automatically.

Worked Example — Equation From Gradient & Point02

Find the equation of the line with gradient 3 passing through (4, −1). [3 marks]

M

Method: use point-slope form y − y₁ = m(x − x₁) with m = 3 and (x₁, y₁) = (4, −1).

C

Calculate: y − (−1) = 3(x − 4) → y + 1 = 3x − 12 → y = 3x − 13.

E

Evaluate: answer y = 3x − 13. Check — at x = 4: y = 12 − 13 = −1 ✓. Gradient is 3 ✓. M1 substitute correctly, M1 expand, A1 final equation in y = mx + c form.

Worked Example — Equation Through Two Points03

Find the equation of the straight line through P(2, 5) and Q(5, 14). [3 marks]

M

Method: (1) find gradient with m = (y₂ − y₁)/(x₂ − x₁); (2) substitute into point-slope with either point.

C

Gradient: m = (14 − 5)/(5 − 2) = 9/3 = 3.

Point-slope with P(2, 5): y − 5 = 3(x − 2) → y − 5 = 3x − 6 → y = 3x − 1.

E

Evaluate: answer y = 3x − 1. Check using Q(5, 14): y = 3(5) − 1 = 14 ✓. Both points satisfy the equation. M1 gradient, M1 substitute point, A1 final equation.

Worked Example — Perpendicular Line

HIGHER

Find the equation of the line perpendicular to y = 2x − 5 that passes through (4, 7). [4 marks]

M

Method: (1) read gradient of given line (m₁ = 2); (2) perpendicular gradient m₂ = −1/m₁ = −½; (3) use point-slope with (4, 7).

C

Perpendicular gradient: m₂ = −1/2.

Point-slope: y − 7 = −½(x − 4) → y − 7 = −x/2 + 2 → y = −x/2 + 9.

E

Evaluate: answer y = −x/2 + 9 (equivalently x + 2y = 18). Check — gradient −½ × original 2 = −1 ✓; at x = 4: y = −2 + 9 = 7 ✓. M1 perpendicular gradient, M1 point-slope setup, M1 expand, A1 final equation.

Examiner Alert: "Flip AND Negate"

⚠️ EXAMINER ALERT: The single most-missed mark on perpendicular-line questions. Students give the perpendicular gradient as just the negative of the original, or just its reciprocal — but not both. AQA mark schemes explicitly reject "same gradient, or negated-but-not-reciprocal" as wrong.

Original gradientWrong (negate only)Wrong (flip only)Correct: flip AND negate

2−2 ✗½ ✗−½ ✓

−33 ✗−⅓ ✗⅓ ✓

¾−¾ ✗4/3 ✗−4/3 ✓

−⅖⅖ ✗−5/2 ✗5/2 ✓

The always-works check

Multiply your perpendicular gradient by the original. If the answer is exactly −1, you're correct. If it's anything else, re-apply the flip-and-negate.

Examiner Alert: Gradient Arithmetic

⚠️ EXAMINER ALERT: On AQA Jun23 Paper 1 Q24, "nearly half lost full marks after making a simple arithmetic slip when computing the gradient — e.g. 22 − 16 = 8". Double-check every subtraction, especially when coordinates include negatives.

Order Matters

If you write y₂ − y₁ on top, you must write x₂ − x₁ on the bottom. Same order, top and bottom.

Swap one: sign flips wrongly.

Negatives Bite

Example: from (−2, 3) to (4, 11).

Run = 4 − (−2) = 6, not 2.

Rise = 11 − 3 = 8.

Gradient = 8/6 = 4/3.

Sanity check

After computing m, look at the line: if it slopes up, m must be positive. If it slopes down, m must be negative. A sign mismatch means you've swapped the order.

The MCE Framework for Line Equations

Every line-equation question — gradient, parallel, perpendicular, or two-point — follows the same three-step structure.

MMethod: Identify what you're given (equation? two points? perpendicular?). Choose the right formula.

CCalculate: Find m first. Then substitute into point-slope. Expand carefully. Watch every sign.

EEvaluate: Check your equation passes through the given point(s). For perpendicular, check m₁ × m₂ = −1.

Marks map directly

A typical 4-mark perpendicular question is marked: M1 perpendicular gradient · M1 substitute point · M1 expand · A1 final equation. Each MCE step earns a mark — showing working survives arithmetic slips.

Practice Question 1

A straight line has equation 2y = 6x − 10.

(a) Write down the gradient of the line. [1 mark]

(b) Write down the coordinates of the point where the line crosses the y-axis. [1 mark]

(c) A second line is parallel to this one and passes through the origin. Write down its equation. [1 mark]

M

Method: rearrange to y = mx + c form first. Then read off m and c. For the parallel line, use the same m with c = 0.

C

Rearrange: 2y = 6x − 10 → divide both sides by 2 → y = 3x − 5.

(a) Gradient m = 3.

(b) y-intercept: c = −5, so the line crosses the y-axis at (0, −5).

(c) Parallel → same gradient 3. Through origin → c = 0. Equation: y = 3x.

E

Evaluate: Check (b) — substitute x = 0 into y = 3x − 5 → y = −5 ✓. Check (c) — gradient matches, passes through (0, 0) ✓.

Practice Question 2

A straight line passes through A(−1, 4) and B(3, 12).

Work out the equation of line AB. Give your answer in the form y = mx + c. [3 marks]

M

Method: (1) find m from the two points; (2) apply point-slope with one of the points; (3) rearrange to y = mx + c.

C

Gradient: m = (12 − 4) / (3 − (−1)) = 8 / 4 = 2.

Point-slope using A(−1, 4): y − 4 = 2(x − (−1)) → y − 4 = 2(x + 1) → y − 4 = 2x + 2 → y = 2x + 6.

E

Evaluate: answer y = 2x + 6. Check with B(3, 12): y = 2(3) + 6 = 12 ✓. M1 gradient computed correctly, M1 substitute point into point-slope, A1 correct equation in y = mx + c form.

Practice Question 3 HIGHER

The line L has equation y = 4x − 3.

A second line is perpendicular to L and passes through the point (8, 5).

Find the equation of the second line in the form y = mx + c. [4 marks]

M

Method: (1) gradient of L is 4; (2) perpendicular gradient is −1/4 (flip AND negate); (3) point-slope with (8, 5); (4) rearrange.

C

Perpendicular gradient: m = −1/4.

Point-slope: y − 5 = −¼(x − 8) → y − 5 = −x/4 + 2 → y = −x/4 + 7.

E

Evaluate: answer y = −x/4 + 7. Perpendicularity check: 4 × (−¼) = −1 ✓. Point check: at x = 8, y = −2 + 7 = 5 ✓. M1 perpendicular gradient, M1 substitute point, M1 expand, A1 correct final equation.

Summary

Key Points

y = mx + c: m is the gradient, c is the y-intercept. Rearrange to this form before reading them off.

Gradient from 2 points: m = (y₂ − y₁)/(x₂ − x₁) — consistent order, top and bottom.

Parallel lines: same m, different c.

Perpendicular lines HIGHER : m₁ × m₂ = −1 → m₂ is the negative reciprocal (flip AND negate).

Point-slope form: y − y₁ = m(x − x₁) — works for every "find the equation" question.

Definitely Do

Rearrange to y = … before reading m and c.

Check the sign of m against the line's slope.

For perpendicular: verify m₁ × m₂ = −1.

Avoid

Just negating (or just flipping) for perpendicular.

Swapping the order of subtraction in the gradient formula.

Arithmetic slips on negatives — always double-check.

Exam Tip

Line-equation questions are highly predictable. Whenever you see a gradient and a point, your hand should already be writing "y − y₁ = m(x − x₁)" — that's M1 before you've even thought about the numbers.