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NotesMath AATopic 2.1Equations of lines
Back to Math AA Topics
2.1.12 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 steep a line is — rise ÷ run.

Picture going right one step: → m = 3 means go up 3 → m = −2 means go down 2 → m = 0 means stay flat To find it exactly, 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

Drag the two points to watch the rise, the run and the gradient value change together. Make the line steeper, flatter, or tilt it downhill to see m turn negative.

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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. Gradient formula — subtract in the same order, top and bottom.
  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.

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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
EquationGradient my-intercept c
y = 3x + 535
y = −x + 2−12
y = ½x − 4½−4 (keep the minus!)
y = 606 (flat line)

Each line here is one row from the table above. Pick an equation: c is where it crosses the y-axis, and m is the rise for each step right. Notice y = 6 is perfectly flat — that is m = 0.

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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.

IB-style question — gradient from ax + by + d = 0

Find the gradient of the line 4x + 3y − 12 = 0.

Step by step

  1. Make y the subject — move the x-term and the constant to the other side.
  2. Divide every term by 3.
  3. The gradient is the number multiplying x.

Final answer

Gradient m = −4/3. Shortcut: straight from ax + by + d = 0, the gradient is m = −a/b = −4/3 — this exact step shows up in real exams.

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Point + gradient, or two points: Given a gradient and a point: put the gradient into y = mx + c, then substitute the point to find c.

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 with gradient 3 that passes through (2, 5).

Give your answer as y = mx + c.

Step by step

  1. Start with y = mx + c, using m = 3.
  2. Substitute the point (2, 5): put x = 2 and y = 5.
  3. Solve for c.
  4. Write the full equation.

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 formula first.
  2. Start y = 3x + c and substitute one point, say (1, 2).
  3. Solve for c.
  4. Write the full equation.

Final answer

y = 3x − 1.

Both worked examples build the same line, y = 3x − 1. Watch it drawn: Step 1 plots the y-intercept (0, −1); Step 2 uses the gradient — right 1, up 3; Step 3 draws the full line.

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Faster alternative — point–gradient form: Another standard method (also in the formula booklet) is point–gradient form: y − y₁ = m(x − x₁).

Drop in the point and gradient, then expand — e.g. y − 5 = 3(x − 2) → y = 3x − 1.

Same answer; use whichever you find clearer.

(With two points, either point works.)
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 Topics

Continue learning with these related topics from the same unit:

2.1.2Parallel lines
2.1.3Perpendicular lines
2.1.4Perpendicular bisector
2.2.1Function notation
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