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Cambridge IGCSE Mathematics — 0580 Extended

Topic 3.5: Coordinate geometry — Equations of Lines

Forms of a straight-line equation

A straight line can be written in several equivalent forms. The form you use depends on what the question gives you — and what it asks for.

FormMeaningExample
\(y = mx + c\)Gradient \(m\), \(y\)-intercept \(c\)\(y = 2x - 3\)
\(ax + by = c\)General linear form (rearrange to find \(m\))\(5x + 4y = 8\)
\(y = k\)Horizontal line (gradient \(0\))\(y = 3\)
\(x = k\)Vertical line (gradient undefined)\(x = -2\)

Method

  1. If you need \(m\) or \(c\), rearrange into \(y = mx + c\).
  2. If you need an equation from a graph or points, find \(m\), then substitute a known point.
  3. Fully simplify: cancel common factors; leave fractions in lowest terms.

Gradient and intercept from an equation

When the equation is not already \(y = mx + c\), rearrange so \(y\) is the subject. Then \(m\) is the coefficient of \(x\) and \(c\) is the constant term.

Rearrange \(5x + 4y = 8\) into the form \(y = mx + c\). State the gradient and the \(y\)-intercept.

Steps rearranging 5x plus 4y equals 8 into y equals negative five-quarters x plus 2
\(m = -\dfrac{5}{4}\), \(c = 2\).

Paper 2 (non-calculator)

Keep the fraction: \(m = -\dfrac{5}{4}\) is preferred to a decimal. If you divide every term, do not forget to divide the constant as well.

Equation from two points

Find the gradient first, then substitute either point into \(y - y_1 = m(x - x_1)\) (or into \(y = mx + c\) to find \(c\)). Simplify fully.

Method

  1. \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
  2. Substitute one point: \(y - y_1 = m(x - x_1)\).
  3. Expand and simplify to \(y = mx + c\) (or another fully simplified form if asked).

Find the equation of the line through \(A(1, 2)\) and \(B(3, 6)\).

Line through A(1, 2) and B(3, 6) with gradient 2 giving equation y equals 2x
\(m = 2\); using \(A\): \(y - 2 = 2(x - 1)\) simplifies to \(y = 2x\).

Gradient and a point; horizontal and vertical

If you know \(m\) and one point, use the same point–gradient step. Horizontal lines are \(y = k\); vertical lines are \(x = k\).

A line has gradient \(\dfrac{1}{2}\) and passes through \(P(2, 1)\). Find its equation.

Line of gradient one half through P(2, 1) giving y equals one half x
\(y - 1 = \dfrac{1}{2}(x - 2)\) simplifies to \(y = \dfrac{1}{2}x\).

Identify the equations of the horizontal and vertical lines shown.

Horizontal line y equals 3 and vertical line x equals negative 2 on Cartesian axes
Horizontal: \(y = 3\). Vertical: \(x = -2\).

Try this

Find the gradient and \(y\)-intercept of \(3x - 6y = 12\). Then find the equation of the line through \((0, 4)\) and \((2, 0)\).

Show answer
Answer

  1. Rearrange \(3x - 6y = 12\)

    \[ -6y = -3x + 12 \Rightarrow y = \dfrac{1}{2}x - 2 \]

  2. So \(m = \dfrac{1}{2}\), \(c = -2\)

    \[ m = \tfrac{1}{2},\quad c = -2 \]

  3. Two points: find \(m\)

    \[ m = \dfrac{0 - 4}{2 - 0} = -2 \]

  4. Through \((0, 4)\): \(c = 4\)

    \[ y = -2x + 4 \]

Exam Traps

  • Do not leave \(y - 2 = 2(x - 1)\) as the final answer unless the question allows it — CIE usually wants a fully simplified equation.
  • A vertical line is \(x = k\), not \(y = k\). Its gradient is undefined — do not write \(m = 0\).

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