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NotesMath AA SLTopic 2.1Equations of lines
Back to Math AA SL Topics
2.1.11 min read

Equations of lines

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Gradient — rise over run
  • The three forms of a line
  • Build a line from what you're given
  • Intercepts — where it crosses the axes
Gradient measures steepness: The gradient m is how much the line rises for each step across — rise ÷ run. Read two points off the line and divide the change in y by the change in x.
one point on the line
another point on the line

IB-style question — gradient from two points

Find the gradient of the line through A(1, 2) and B(4, 11).

Step by step

  1. Subtract the y-values (top) and the x-values (bottom) in the same order.
  2. Work it out.

Final answer

m = 3 (the line rises 3 for every 1 across).

What the sign tells you

  • m > 0 → uphill (left to right).
  • m < 0 → downhill.
  • m = 0 → horizontal line y = c.

Watch out

  • Subtract in the same order top and bottom.
  • A vertical line x = a has no gradient (run = 0).
  • rise/run, never run/rise.
Same line, three outfits: A straight line can be written three ways — pick whichever fits the question.

Gradient–intercept

  • m = gradient
  • c = y-intercept
  • Best for graphing

Point–gradient

  • Use a point + gradient
  • Best for building a line

General form

  • Tidy integer form
  • Gradient = −a/b

IB-style question — switch between forms

Write y − 3 = 2(x − 1) in the form y = mx + c, then in the form ax + by + d = 0.

Step by step

  1. Expand the bracket.
  2. Make y the subject → gradient–intercept form.
  3. Move everything to one side → general form.

Final answer

y = 2x + 1, or equivalently 2x − y + 1 = 0.

Gradient straight from ax + by + d = 0: Rearrange to y = mx + c: the gradient is m = −a/b. (e.g. 3x + 2y − 6 = 0 → y = −1.5x + 3, so m = −1.5.) This exact step shows up in real exams.

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Point + gradient, or two points: Given a gradient and a point, drop them straight into point–gradient form. Given two points, find the gradient first, then do the same.

IB-style question — a point and a gradient

Find the equation of the line through (2, 5) with gradient 3. Give your answer as y = mx + c.

Step by step

  1. Point–gradient form with (x₁, y₁) = (2, 5), m = 3.
  2. Expand and tidy.

Final answer

y = 3x − 1.

IB-style question — two points

Find the equation of the line through P(1, 2) and Q(3, 8).

Step by step

  1. Gradient first.
  2. Use point–gradient with either point, say (1, 2).
  3. Tidy.

Final answer

y = 3x − 1.

Either point works: With two points you can substitute either one into point–gradient form — you get the same line. Pick the one with smaller numbers.
Set the other coordinate to zero: y-intercept: set x = 0. x-intercept: set y = 0 and solve. (And in y = mx + c, the number c is the y-intercept — read it straight off.)

IB-style question — both intercepts

Find where the line y = 2x − 6 crosses each axis.

Step by step

  1. y-intercept: put x = 0.
  2. x-intercept: put y = 0 and solve.

Final answer

Crosses the y-axis at (0, −6) and the x-axis at (3, 0).

Don't swap the coordinates: The y-intercept is the point (0, c) and the x-intercept is (x, 0) — the zero goes in different slots. A common slip is writing (−6, 0) for the y-intercept.

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Find the gradient of the line passing through A(−1, 3) and B(2, 12). [2 marks]

Related Math AA SL Topics

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2.1.2Parallel & perpendicular
2.2.1Function notation
2.2.2Domain & range
2.2.3Inverse as reflection
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