Graphs: use the formy = mx + c to identify parallel lines find the equation of the line through two given points, or through one point with a given gradient

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Overview

A straight-line equation is a compact set of instructions for every point on a graph. In y = mx + c, the gradient m tells you how y changes when x increases, while c tells you where the line crosses the y-axis. This guide develops three dependable exam skills: recognising parallel lines, finding an equation through two points, and finding an equation through one point when the gradient is known.

These skills combine coordinates, substitution, negative numbers and rearranging equations. They also support simultaneous equations, linear sequences and interpreting real-life rates. Candidates may be asked to write down a gradient, explain why lines are parallel, work out an equation or interpret a graph in context. The method matters because visible, correct working can receive credit even when a later calculation goes wrong, subject to the question's mark scheme.

The relevant reference is AQA GCSE Mathematics 8300, 3.2.2 Graphs, A9, supported by A8 and A10. This topic is assessed on both Foundation and Higher tiers. Identifying perpendicular lines is a separate Higher-only extension. 1

Key Concepts

1. Understand the roles of m and c

For y = 2x + 3, m = 2 and c = 3. The line crosses the y-axis at (0, 3), not (3, 0). This follows directly from substitution: at the y-axis, x = 0, so y = 2(0) + 3 = 3. The gradient is not the height of the line. It describes a rate of change.

Increase x from 1 to 2. The corresponding y-values change from 5 to 7, an increase of 2. Every one-unit increase in x produces the same two-unit increase in y. That constant rate makes the graph straight. In y = -2x + 3, the gradient is -2 and the intercept is still +3. Read the sign with its number.

A fractional gradient is equally meaningful. For m = 3/2, moving two units right moves three units up. Do not confuse m = 3/2 with 2/3. Gradient compares vertical change with horizontal change, in that order. A larger absolute gradient means a steeper line when graphs use the same axis scales.

Cover and recall: what remains of y = mx + c when x = 0? Explain why that identifies the intercept without drawing anything.

2. Calculate gradient from two points

For points (x₁, y₁) and (x₂, y₂), calculate m = (y₂ - y₁)/(x₂ - x₁). The numerator is the change in y; the denominator is the corresponding change in x. Keep the subtraction order consistent.

Through (1, 5) and (3, 11), m = (11 - 5)/(3 - 1) = 6/2 = 3. Reversing both subtractions gives (5 - 11)/(1 - 3) = -6/-2 = 3, the same result. Reversing only one produces -3 incorrectly. Write the substituted fraction before simplifying so the examiner can see the method used.

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The dashed triangle shows a run of 2 and rise of 6. Its sloping side is not used in the gradient calculation. On a graph, choose two accurately readable, well-separated points on the line. A larger triangle reduces the effect of small reading errors. Count coordinate units, not simply squares, if the axes have different scales.

Why does this work? Subtract y₁ = mx₁ + c from y₂ = mx₂ + c. The intercept cancels, leaving y₂ - y₁ = m(x₂ - x₁). Dividing by the non-zero horizontal change gives the gradient formula.

3. Find the equation through two points

Use GSC: Gradient, Substitute, Check. First calculate m. Then put one complete coordinate pair into y = mx + c and solve for c. Finally write the equation and test the other point.

Through (1, 5) and (3, 11), the gradient is 3. Substitute (1, 5): 5 = 3(1) + c, giving c = 2. Therefore y = 3x + 2. Check (3, 11): 3(3) + 2 = 11. Finding m = 3 and c = 2 without writing an equation does not fully answer 'Work out the equation'.

For negative coordinates, preserve brackets. Through (-2, 5) and (4, 2), m = (2 - 5)/(4 - (-2)) = -3/6 = -1/2. Using (4, 2), 2 = (-1/2)(4) + c, so c = 4. The equation is y = -x/2 + 4. Checking (-2, 5) gives -(-2)/2 + 4 = 5.

Keep exact fractions unless rounding is requested. Decimal 0.5 is exact, but 0.33 is not exactly 1/3. Premature rounding can change the intercept and lose accuracy. Candidates are credited for valid alternative algebraic methods; this guide's suggested mark allocations are not universal AQA rules. 4

4. Find the equation through one point with a known gradient

When m is given, skip the gradient calculation. For a line of gradient 3 through (2, -1), begin y = 3x + c. Substitute the coordinates in their correct positions: -1 = 3(2) + c. Therefore c = -7, and y = 3x - 7. Check: 3(2) - 7 = -1.

A known point is not automatically the intercept. The point (2, -1) lies on the line, but its x-coordinate is not zero. Writing c = -1 mistakes an ordinary y-coordinate for the y-intercept. If the given point is (0, -1), however, c really is -1.

The equivalent point-gradient form is y - y₁ = m(x - x₁). For this example, y + 1 = 3(x - 2), which expands to y = 3x - 7. This is a useful alternative, not a requirement to learn a separate method. Return the form requested in the question.

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5. Identify parallel lines, including disguised equations

Two distinct non-vertical straight lines are parallel when their gradients match. Thus y = 2x + 3 and y = 2x - 1 are parallel: both have m = 2, but their intercepts differ. Equal intercepts alone do not establish parallelism. For example, y = 2x + 3 and y = -x + 3 meet at (0, 3).

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The vertical gap between the two illustrated lines is always 4 because (2x + 3) - (2x - 1) = 4. The intercept changes the starting height, not the rate. If both m and c match, the equations represent the same, coincident line, rather than two distinct parallel lines.

Always make y the subject before comparing coefficients. For 2y = 6x - 4, divide every term by 2: y = 3x - 2. Its gradient is 3, not 6. Likewise, 3x + 2y = 8 becomes 2y = -3x + 8, then y = -3x/2 + 4. The gradient is -3/2.

To find a parallel line through a specified point, copy the gradient and find a new intercept. A line parallel to 2y = 6x - 4 through (1, 7) has y = 3x + c. Then 7 = 3 + c gives y = 3x + 4. Explain parallelism using equal gradients, not appearance alone.

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6. Handle horizontal and vertical lines correctly

A horizontal line such as y = 4 has gradient 0: y does not change as x changes. It fits y = mx + c with m = 0 and c = 4. Distinct horizontal lines are parallel.

A vertical line such as x = 4 has an undefined gradient, not zero. The horizontal change is zero, so the gradient fraction would divide by zero. It cannot be expressed as y = mx + c with finite m. Through (2, 1) and (2, 7), the line is x = 2. Distinct vertical lines are parallel to one another. A horizontal and a vertical line are perpendicular, not parallel.

Higher-only extension: for non-vertical, non-horizontal perpendicular lines, m₁m₂ = -1, so one gradient is the negative reciprocal of the other. A gradient of 2 gives a perpendicular gradient of -1/2. This is not the rule for parallel lines and is beyond the requested core. 1

Mathematical Relationships and Formula-Sheet Status

The published 2027 Foundation and Higher formula sheets do not print the straight-line formulas below. 'Must memorise' here means they are not supplied there; simple rearrangements can be derived from the core equation rather than learnt separately. Check the sheet for your examination year. 3 5 6

Must memorise: y = mx + c. x and y are coordinates, m is gradient and c is the y-intercept value. Use for non-vertical lines.

Must memorise: m = Δy/Δx = (y₂ - y₁)/(x₂ - x₁). Δ means change. Use two distinct points with x₂ ≠ x₁, or a gradient triangle.

Must memorise, or derive: c = y₁ - mx₁ and y - y₁ = m(x - x₁). Both follow from substituting a known point. The intercept point is (0, c).

Must memorise: m₁ = m₂ for parallel non-vertical lines. For distinct lines, c₁ ≠ c₂. Matching both values means coincident lines.

Must memorise, or derive: y = k is horizontal with m = 0; x = k is vertical with undefined gradient. Here k is a fixed constant.

Must memorise, or derive: x = -c/m for the x-intercept when m ≠ 0. Set y = 0. A horizontal line y = c ≠ 0 has no x-intercept; y = 0 is the x-axis itself.

Higher-only, must memorise: m₁m₂ = -1, equivalently m₂ = -1/m₁, for the perpendicular-gradient extension where these gradients are defined and non-zero.

Graph Skills: Plot, Read and Interpret

For y = 2x - 1, use x = -1, 0 and 2 to obtain y = -3, -1 and 3. Plot (-1, -3), (0, -1) and (2, 3), then draw one ruled straight line over the required interval. Two points determine a line; a third helps detect a substitution error. Label axes, read scales and distinguish a sketch from an accurate plot.

To recover an equation from a graph, read the y-intercept, calculate the gradient from two clear points and combine them. For negative gradients, the rise is negative when moving right. If the intercept lies outside the visible grid, calculate m from visible points and substitute one to find c. Do not guess an unseen crossing.

In AQA June 2024 8300/3H question 7, the three-mark scheme credits two correct points, plotting the candidate's points and the correct straight line across the required interval. This is a genuine example of method and accuracy marks, not a fixed allocation for all graph questions. 4

Practical Applications and Units

A taxi model C = 2d + 4 has gradient £2 per kilometre and a £4 starting charge. A second model C = 2d + 7 is parallel: the same per-kilometre charge, but £3 more for every distance. State what the gradient and intercept mean in the context, not just their numerical values.

On a distance-time graph, gradient has distance units divided by time units. A rise of 120 metres in 2 minutes is 60 m/min, or 1 m/s. Convert 2 minutes to 120 seconds before calculating in m/s. For cost, 250 pence is £2.50, so mixing pounds and pence changes the numerical gradient by a factor of 100. Pure coordinate questions need no invented physical units.

There is no required practical for this mathematics topic. These applications develop modelling and interpretation, not a science practical method. For costs or travel, consider the meaningful domain: negative distance may be mathematically possible but irrelevant to the real model.

Listen, Retrieve and Revisit

The ten-minute female-tutor podcast follows introduction, core concepts, exam tips, recall quiz and summary. Pause before quiz answers and calculate rather than just listening. Afterwards, cover the guide and write the complete GSC method from memory. Solve one equation question now, another tomorrow and a mixed question several days later.

Elaborate: why does subtracting the two point equations remove c? What changes if c increases but m stays fixed? What if both points have the same x-coordinate? Explain each answer using the mathematics. This links retrieval to understanding and helps distinguish similar-looking exam questions.

References