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GCSE/Mathematics/AQA

A9Plot linear graphs; y = mx + c; parallel and perpendicular lines

Notes

Linear graphs and the form y = mx + c

Every straight-line graph can be written y = mx + c, where m is the gradient (steepness) and c is the y-intercept (where the line crosses the y-axis). Master this form and you can sketch, analyse, and find equations of straight lines fluently.

Reading a line from its equation

In y = 3x - 4:

  • gradient m = 3 (line rises 3 for every 1 right)
  • y-intercept c = -4 (line crosses the y-axis at (0, -4))

To plot, mark (0, -4), then go right 1 and up 3 to (1, -1), repeat to reach more points. Join with a ruler.

Finding the equation from a graph

  1. Read c directly: where does the line meet the y-axis?
  2. Pick two clear lattice points and compute m = (y₂ − y₁)/(x₂ − x₁).
  3. Write y = mx + c.

If a line passes through (0, 5) and (4, 13): c = 5, m = (13 − 5)/(4 − 0) = 2, so y = 2x + 5.

Equation given a point and gradient

Use point-gradient form, then rearrange to y = mx + c: y - y₁ = m(x - x₁).

Example: line through (3, 1) with gradient 2. y - 1 = 2(x - 3) ⇒ y = 2x - 5.

Equation through two points

Compute m, then use one point. From (1, 2) and (4, 11): m = 9/3 = 3. Use (1, 2): y - 2 = 3(x - 1) ⇒ y = 3x - 1.

Parallel and perpendicular lines

  • Parallel lines have the same gradient: y = 3x + 7 and y = 3x − 5 are parallel.
  • Perpendicular gradients multiply to −1: if one line has gradient 2, a perpendicular line has gradient −1/2. Symbolically, m₁ × m₂ = −1 so m₂ = −1/m₁ (the negative reciprocal).

Worked example: find the equation of the line through (4, 1) perpendicular to y = 2x + 7. The given gradient is 2. Perpendicular gradient = −1/2. Use point-gradient: y − 1 = −½(x − 4) ⇒ y = −½x + 3.

Special cases

  • Horizontal line: y = c (gradient 0).
  • Vertical line: x = a (gradient undefined; not in the form y = mx + c).

Common mistakesCommon mistakes (examiner traps)

  1. Confusing gradient and intercept. In y = mx + c, m is the coefficient of x, c is the number on its own. Read carefully.
  2. Forgetting to rearrange. A line written as 2y = 4x − 6 is not in y = mx + c form yet. Divide by 2 first: y = 2x − 3.
  3. Reciprocal vs negative reciprocal. Perpendicular gradient is the negative reciprocal: 3 → −1/3, NOT 1/3.
  4. Sign of c when reading the y-intercept. If the line crosses the y-axis below the origin, c is negative.
  5. Picking unclear points when reading m off a graph. Always use lattice (whole-number) points.

Try thisQuick check

Find the equation of the line through (-1, 4) parallel to y = -2x + 5. Gradient is -2 (parallel). y - 4 = -2(x + 1) ⇒ y = -2x + 2.

AI-generated · claude-opus-4-7 · v3-deep-algebra

Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    Identify gradient and intercept

    (F1) Write down the gradient and y-intercept of y = -3x + 7.

    [Foundation tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  2. Question 23 marks

    Plot a line from its equation

    (F2) Plot the line y = 2x - 1 for x = -2 to x = 3 by completing a table of values, then draw the line.

    [Foundation tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  3. Question 32 marks

    Equation from gradient + intercept

    (F3) A line has gradient 4 and crosses the y-axis at -3. Write its equation.

    [Foundation tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  4. Question 43 marks

    Equation through two points

    (F/H4) Find the equation of the line through (2, 7) and (5, 16).

    [Foundation/Higher crossover]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  5. Question 53 marks

    Rearrange before reading m and c

    (F/H5) Find the gradient and y-intercept of the line 3y - 6x = 12.

    [Foundation/Higher crossover]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  6. Question 63 marks

    Parallel line through a point

    (H6) Find the equation of the line parallel to y = -½x + 6 passing through (4, -1).

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  7. Question 73 marks

    Perpendicular through a point

    (H7) L₁: y = 3x - 4. Find the equation of the line L₂ perpendicular to L₁ passing through (6, 5).

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

Flashcards

A9 — Linear graphs

10-card SR deck for AQA GCSE Maths topic A9

10 cards · spaced repetition (SM-2)