Chapter 4 · Geometry and Equations · Concept 2
4-8 Equation of a Straight Line
A line is an endless collection of points, and an equation is a rule that says which points belong to it. The Point! box lists seven cases — but there are really only two ideas in there.
Guiding question
What is the smallest amount you need to know to pin a line down?
Learning outcome. Find the equation of a straight line in its different forms.
Two ideas, written several ways
Idea 1 — lines with no slope worth naming
· through (x₁, y₁) parallel to the x-axis → y = y₁
· through (x₁, y₁) perpendicular to the x-axis → x = x₁
· through (x₁, y₁) parallel to the y-axis → x = x₁
· through two points with the same y → y = y₁
· through two points with the same x → x = x₁
Five bullets, one idea: if the line is flat its equation fixes y, and if it is upright its equation fixes x. Notice that “parallel to the y-axis” and “perpendicular to the x-axis” are the same instruction said twice.
Idea 2 — everything else
y − y₁ = a(x − x₁)
where a is the gradient. If you are given two points instead, work the gradient out first:
a = y₂ − y₁x₂ − x₁
The one that gets written backwards.
“Parallel to the x-axis” gives y = …, not x. The instruction names the axis the line COPIES, and the equation fixes the other letter. When in doubt, draw it — a flat line through (3, −4) obviously has every point at height −4.
Pick a case and draw it. Once you have seen a flat line through (3, −4), you will never write x = 3 for it again.
the line through (3, −4), parallel to the x-axis
(f) Find the equation of the line through (1, −1) and (4, 0).
- 1
Check first whether it is one of the easy kinds.
the x's are 1 and 4 — different. the y's are −1 and 0 — different.If either pair matched, the answer would be one line long. They do not, so there is a gradient to find.
Work with a partner. For each one, say out loud which of the two ideas it needs before writing anything. Half of these do not need a gradient at all, and spotting that is the fastest mark on the page.
Question 6 has a negative gradient.
(1 − (−6)) ÷ (−4 − (−2)) = 7 ÷ (−2) = −7/2. Two subtractions of negatives in one line — work the top and the bottom out separately before dividing.
Fifteen of these twenty need no gradient.
(a) to (o) are all flat or upright lines. Only (p) to (t) use y − y₁ = a(x − x₁). Sorting them into the two kinds first turns a twenty-part question into a five-part one.
An environmental team is monitoring sea level in a coastal estuary. The x-axis is the distance from a research station and the y-axis is the local sea-surface height. Two sensors record the same steady height: Sensor West at A(−4, 2) and Sensor East at B(9, 2). Find the equation of the straight-line boundary joining them.
Writing x = 3 for a line parallel to the x-axis.
Parallel to the x-axis means flat, so the equation fixes the height: y = something. The named axis is the one the line copies, not the letter in the answer.
Hunting for the gradient of an upright line.
Through (−3, 1) and (−3, 2) the change in x is 0, so the gradient does not exist. The answer is x = −3, and no division is needed.
Losing a sign in the gradient.
(1 − (−6)) ÷ (−4 − (−2)) is 7 ÷ (−2). Work out the top and the bottom separately before dividing.
Substituting into y − y₁ = a(x − x₁) with the signs unflipped.
For the point (−1, 5), it is y − 5 = a(x + 1). The minus in the formula turns the −1 into a +1.
Not checking with the second point.
When two points were given, put the OTHER one into your finished equation. It takes five seconds and it catches every arithmetic slip.
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