01 — Explanation
What does each part of y = mx + c tell me?
is the gradient (steepness/direction), and is the -intercept — the value of where the line crosses the -axis (i.e. where ). Once you know and , you know the entire line.
02 — Explanation
How do I find the equation when I can see the graph?
Read straight off the graph (where it crosses the -axis), then find using rise over run between two clearly-plotted points (see E3.3). Substitute both into .
03 — Explanation
How do I find the equation of a line if I'm only given two coordinates, with no graph at all?
This combines two skills you already have: use the gradient formula from E3.3 to calculate from the two points, then substitute and one of the points into and solve for . This method works even when there's nothing to read off a diagram.
04 — Explanation
How do I rearrange other forms of a line into y = mx + c?
Lines are sometimes given as or simply . For :
A line of the form is a vertical line — it has no gradient in the sense and cannot be written that way at all.
05 — Explanation
How do I give my final answer in "fully simplified form"?
Cambridge expects fractions cancelled and no unnecessary decimals — e.g. write rather than unless a decimal is specifically requested.
Exam tip
"Find the equation of the line" always means find the whole expression with actual numbers substituted for and — not just stating the gradient. Missing off the term is a very easy mark to drop.
06 — Explanation
How do I check my final equation is actually correct?
Substitute both of the original coordinates back into your equation — if a point genuinely lies on the line, its and values must satisfy exactly. This costs seconds but catches almost every arithmetic slip, particularly sign errors made while solving for , and is well worth doing on any "find the equation" question worth more than one mark.
07 — Worked example
Worked example
Find the gradient and -intercept of the line with equation .
- Rearrange to make the subject: .
- Divide every term by : , i.e. .
- Compare with : gradient , -intercept .
08 — Worked example
Worked example: finding the equation from two points
Find the equation of the line passing through and .
- Find the gradient: .
- Substitute the gradient and one point, say , into : .
- The equation of the line is .
- Check with the other point: gives ✓, confirming the equation is correct.