Equation of a Line

One form does almost all the work in 0606:

$y - y_1 = m(x - x_1)$

A gradient and any point go straight in, no intercept-solving step, no simultaneous fiddling. The alternatives ($y = mx + c$, $ax + by = c$) are output formats, not working tools: build in point-gradient form, rearrange at the end if asked.

The two standard builds

From a point and a gradient: direct substitution.

Line through $(2, -5)$ with gradient $3$: $y + 5 = 3(x - 2) \to y = 3x - 11$

From two points: gradient first, then either point.

Through $(1, 4)$ and $(3, 10)$: $m = \tfrac{6}{2} = 3 \to y - 4 = 3(x - 1) \to y = 3x + 1$ Self-check: does the other point satisfy it? $10 = 3(3) + 1$ ✓, ten seconds, catches everything.

The gradient line and the point-gradient line are each method marks; the check is free insurance.

Matching the demanded format

0606 frequently specifies the form: “give your answer in the form $ax + by = c$, where $a$, $b$, $c$ are integers”. That’s a format command, the final accuracy mark attaches to it:

$y = \tfrac{3}{4}x + 2 \to 4y = 3x + 8 \to$ $3x - 4y = -8$ (or equivalently $-3x + 4y = 8$)

Clear fractions, collect, and present with integer coefficients. A correct line in the wrong form drops the mark.

Where line equations come from in practice

The gradient is usually derived, not given: from a parallel/perpendicular condition, from differentiation at a point (tangents and normals), or from a midpoint construction (perpendicular bisector). The build step is identical in all of them, which is why this little routine quietly underwrites a dozen question types across both papers.

Special cases worth ten seconds of memory: horizontal lines are $y = k$ (gradient $0$); vertical lines are $x = k$ (gradient undefined, and not expressible as $y = mx + c$).

Common mistakes

• Signs mangled substituting negative coordinates into $y - y_1 = m(x - x_1)$
• The format instruction ignored (fractions left in an “integer coefficients” answer)
• Vertical lines forced into $y = mx + c$
• Two-point builds never checked against the second point
• $c$ found by substitution into $y = mx + c$ when point-gradient form needed no $c$ at all (slower, more slips)

Full topic context: Straight-Line Graphs notes.