Equation of a straight line

Through a point, parallel or perpendicular, from a graph

In 8 of 20 recent exams
Paper 1 or 2
1 to 3 marks

How it is set in the exam

Paper 1 (no calculator) mostly (14 of the 17 parts set since 2021), 1–3 marks a part. Either a short Section A question of 5–7 marks (question 1 in three versions), often split into the midpoint, the gradient and the line; or one or two parts of a longer Section B question, where the line goes with a parabola or into a chain of functions. On Paper 2 (calculator) it is a short chain inside a longer question: the gradient of a given line, then the perpendicular through a given point, with the GDC finding where a line meets the curve. Common (8 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (6)

“find the equation of the line $L$ that passes through $P$ and is perpendicular to $[AB]$ ”
“find the equation of the perpendicular bisector of $[AB]$ ”
“write down the gradient of $L$ ”
“the graph of the linear function $g$ is shown; write down $g(x)$ in the form $mx + c$ ”
“give your answer in the form $ax + by + d = 0$ , where $a, b, d \in \mathbb{Z}$ ”
“the point $(k, 5)$ lies on $L$ ; find $k$ ”

In disguise (4)

" $h$ is a linear function with $h(1) = 3$ and $h(4) = 9$ ": two values of $h$ are two points on its graph. It often sits inside a longer chain of function parts.
a line through two points of a curve: evaluate the curve at the given $x$ -values first
a gradient from an angle: a line that makes an angle of $30°$ with the positive $x$ -axis (measured anticlockwise) has gradient $\tan 30° = \frac{\sqrt{3}}{3}$ ; a line that falls has $m < 0$
a line that is one side of a triangle or a kite, followed by a vertex found on the line, then an area (Area of a triangle: ½ab sin C, forwards and backwards; a kite is two triangles along a diagonal)

Variants (5)

given: two points, a graph, a line and a point, or a description in words
a perpendicular bisector: the midpoint and the perpendicular gradient are often asked first, as separate parts
a line through the origin ( $c = 0$ )
fraction or surd gradients
follow-ups: a point on the line with one unknown coordinate; an area with a vertex on the line

Looks like it, but isn't (5)

the gradient comes from a derivative, at a point of a curve: Tangent or normal at a given point
a tangent to a curve that is parallel or perpendicular to a given line, or at a given angle to the $x$ -axis, at a point not yet known: Find the point of contact
whether a line meets a parabola, or the $k$ that makes it touch: Two graphs
whether a point is above or below a curve: evaluate and compare (Points on a graph)
a line fitted to a data set on the GDC ("the regression line of $y$ on $x$ "), not a line through exact points: Regression on the GDC

Worked example

Read the line off a graph, then the perpendicular in general form

Paper 1 (no calculator)

Question

The diagram shows the graph of the linear function $g$ . The line crosses the $y$ -axis at $(0, -4)$ and passes through the grid point $(6, 5)$ .
(a) Write down $g(x)$ in the form $mx + c$ . [2]
(b) The line $L$ is perpendicular to the graph of $g$ and passes through the point $Q(1, 4)$ . Find the equation of $L$ , giving your answer in the form $ax + by + d = 0$ , where $a, b, d \in \mathbb{Z}$ . [4]

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See it move: Turn the slope triangle a quarter turn

$g$ through $(0, -4)$ and $(6, 5)$ : $m = \frac{5 - (-4)}{6 - 0} = \frac{9}{6} = \frac{3}{2}$ , and it crosses the $y$ -axis at $-4$ : $y = \frac{3}{2}x - 4$ .
Perpendicular through $Q(1, 4)$ : the slope triangle turns a quarter turn, run 2 and rise 3 become run 3 and rise −2. So $m = -1 \div \frac{3}{2} = -\frac{2}{3}$ : flipped and with the sign changed, and the two gradients multiply to $-1$ .
$y - 4 = -\frac{2}{3}(x - 1)$ ; clear the fractions and collect: $2x + 3y - 14 = 0$ , with whole numbers.
Check: $2(1) + 3(4) - 14 = 0$ , so $Q$ is on it. It meets $g$ at $\left(4, 2\right)$ .
(a) Read off the graph: $c = -4$ at once, and $m = \frac{9}{6} = \frac{3}{2}$ , so $g(x) = \frac{3}{2}x - 4$ . (b) $L$ only borrows the gradient: $-\frac{2}{3}$ through $Q(1, 4)$ gives $2x + 3y - 14 = 0$ .

Steps

1
Get the gradient.
- two points: $m = \frac{y_2 - y_1}{x_2 - x_1}$ . On a graph, choose two points where the line crosses grid lines exactly.
- a given line: rearrange to $y = mx + c$ and read $m$ .
- parallel: the same $m$ . Perpendicular: $-\frac{1}{m}$ .
Get a point on the line:
the given point, the midpoint (for a perpendicular bisector), or a point of a curve (evaluate the curve, or on find the meeting point on the GDC).
Write
$y - y_1 = m(x - x_1)$ . Substituting the point into $y = mx + c$ to find $c$ scores too.
Rearrange to the form asked for,
as an equation. For $ax + by + d = 0$ with $a, b, d \in \mathbb{Z}$ , clear every fraction.
Check
that each point the line should pass through satisfies your equation. For a follow-up point, substitute its known coordinate into the line and solve.

Common mistakes

Naming the line instead of writing an equation
" $L = -\frac{2}{3}x + \frac{14}{3}$ " scores . Write $y = \ldots$ .
Half a negative reciprocal
flipping $\frac{3}{2}$ to $\frac{2}{3}$ without changing the sign, or changing the sign without flipping. The two gradients must multiply to $-1$ .
Reading $m$ straight off $ax + by + d = 0$
the gradient of $2x - 3y + 6 = 0$ is $\frac{2}{3}$ , not $2$ . Rearrange to $y = \ldots$ first.

More mistakes (3)

The right gradient through the wrong point
the line through $Q$ uses $Q$ , and a perpendicular bisector uses the midpoint, not $A$ or $B$ .
Stopping before the form asked for
fractions left in $ax + by + d = 0$ when $a, b, d \in \mathbb{Z}$ lose the last mark.
Rounding a GDC point before using it
keep $A$ at full accuracy to find $c$ , and round only the final answer.

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