Are perpendicular lines on the Foundation paper?

No. Spec point A9 asks both tiers to use y = mx + c to identify parallel lines, but the perpendicular part of that spec point is Higher only, and so is the perpendicular bisector work that goes with it. Foundation candidates still meet parallel lines and finding the equation of a line through two given points.

Do I get the perpendicular gradient rule on the formulae sheet?

No. The Exam Aid sheet issued with each 1MA1 paper carries the area of a trapezium, volume of a prism, the circle formulae, the quadratic formula, Pythagoras' theorem and the trigonometric ratios, the sine and cosine rules, the area of a triangle, compound interest and two probability rules. Nothing on it concerns gradients, so m_1 m_2 = -1 and the gradient formula both have to be recalled.

How can I tell whether two lines are perpendicular from their equations?

Rearrange both into y = mx + c and multiply the two gradients. A product of -1 means perpendicular, and equal gradients mean parallel. For example 3x + y = 5 has gradient -3 and x - 3y = 6 has gradient (1)/(3), and -3 (1)/(3) = -1, so those two are perpendicular.

What is the perpendicular bisector of a line segment?

It is the line that crosses the segment at right angles through its midpoint, so every point on it is the same distance from both ends. To find its equation, work out the midpoint, work out the gradient of the segment, take the negative reciprocal for the bisector's gradient, then substitute the midpoint to find c.

Parallel and Perpendicular Lines - Edexcel GCSE Higher Maths

Typically 3-5 marks per paper, usually one question worth 3-4 · Spec 1MA1

Two lines are parallel when their gradients are equal, and perpendicular when their gradients multiply to $-1$ . Both live in spec point A9 of Edexcel's 1MA1 specification, but they are not assessed the same way: identifying parallel lines from $y = mx + c$ is crossover content that Foundation candidates meet too, while the perpendicular half is Higher only. That single rule, $m_1 m_2 = -1$ , is behind almost everything on this page, including the perpendicular bisector of a line segment.

Edexcel rarely hands you the line in a convenient form. A typical Higher question gives $L$ as $6y + kx = 18$ or $3x + 2y = 7$ and asks for a line perpendicular to it through a stated point, worth 3-4 marks. The rearrangement is where the first mark lives, and it is also where most candidates lose it: taking the coefficient of $x$ straight off $4x + 3y = 15$ gives 4, when the gradient is $-\frac{4}{3}$ . Paper 1 is non-calculator, so a perpendicular gradient of $\frac{3}{4}$ has to be handled as a fraction all the way through the substitution rather than typed in as 0.75.

The mark scheme for these questions is a chain: a process mark for making $y$ the subject, a second for using $m_1 m_2 = -1$ (usually not dependent on the first, so a candidate who rearranges wrongly can still show the negative reciprocal and score), a third for substituting the given point to find $c$ , then the accuracy mark. Half a negative reciprocal earns nothing. Writing $\frac{1}{2}$ as the perpendicular gradient of a line with gradient 2 flips the sign but not the fraction, and both errors turn up on scripts in roughly equal numbers. Practise reading the answer form too, since 'in the form $ax + by = c$ where $a$ , $b$ and $c$ are integers' means one more line of working after you have the equation.

Worked example

The line

L

has equation

4x + 3y = 15

. The line

M

is perpendicular to

L

and passes through the point

(8, 1)

. Find an equation of

M

in the form

ax + by + c = 0

, where

a

b

and

c

are integers.

[5 marks]

Rearrange $L$ to make $y$ the subject: $3y = -4x + 15$ , so $y = -\frac{4}{3}x + 5$ .
P1: a process to make $y$ the subject; moving only the constant term is not enough for this mark
The gradient of $L$ is $-\frac{4}{3}$ , so the gradient of $M$ is $\frac{3}{4}$ , because $-\frac{4}{3} \times \frac{3}{4} = -1$ .
P1: using $m_1 m_2 = -1$ ; turn the fraction upside down and change the sign, both together
Substitute $(8, 1)$ into $y = \frac{3}{4}x + c$ : $1 = \frac{3}{4} \times 8 + c$ , so $1 = 6 + c$ and $c = -5$ .
P1: substituting the given point into $y = mx + c$ with their perpendicular gradient
So $M$ has equation $y = \frac{3}{4}x - 5$ .
A1: a correct equation of $M$ ; check it by putting $x = 8$ back in
Multiply through by 4 and collect on one side: $4y = 3x - 20$ , giving $3x - 4y - 20 = 0$ .
B1: the stated form, with integer coefficients and nothing left as a fraction

Practice questions

These are original questions in Edexcel 1MA1 Higher style. Plenty of them hand you the line as $ax + by = c$ on purpose, because that rearrangement carries the first process mark on the real papers. Work them without a calculator where you can, the way Paper 1 sets them, and type fractional answers directly (-3/4 and -0.75 are both accepted). Questions asking for an equation want the gradient and the intercept as two separate values, so finish the substitution rather than stopping at the gradient.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [1 mark]

Work out the gradient of a line that is perpendicular to the line with equation $y = -4x + 1$ .

Your answer (fractions like 1/2 are fine)
Question 2 Exam pace [3 marks]

The line with equation $2y = kx + 5$ is parallel to the line with equation $3x + y = 9$ . Work out the value of $k$ .

Your answer (fractions like 1/2 are fine)
Question 3 Stretch [4 marks]

The line $L$ passes through the points $A(-1, 2)$ and $B(3, 10)$ . The line $M$ is perpendicular to $L$ and passes through $B$ . Find an equation of $M$ in the form $y = mx + c$ by stating $m$ and $c$ .

m = c =
Question 4 Stretch [4 marks]

The point $A$ has coordinates $(1, 2)$ and the point $B$ has coordinates $(7, 10)$ . Find an equation of the perpendicular bisector of $AB$ in the form $y = mx + c$ by stating $m$ and $c$ .

m = c =
Question 5 Stretch [4 marks]

Two mobile phone masts stand at the points $A(2, 3)$ and $B(10, 7)$ on a map where 1 unit represents 1 km. A boundary line is drawn so that every point on it is the same distance from both masts. Find an equation of the boundary in the form $y = mx + c$ by stating $m$ and $c$ .

m = c =
Question 6 Stretch [4 marks]

The line $L$ has equation $y = 2x - 6$ . The line $M$ is perpendicular to $L$ and passes through the point where $L$ crosses the $x$ -axis. Find an equation of $M$ in the form $y = mx + c$ by stating $m$ and $c$ .

m = c =

Common mistakes examiners see

Reading a gradient straight off $ax + by = c$ without rearranging, so the gradient of $2x + 5y = 20$ is given as 2 and the perpendicular gradient as $-\frac{1}{2}$ instead of $\frac{5}{2}$ .

Make $y$ the subject before you write down any gradient. Divide every term by the coefficient of $y$ , including the constant, and only then read off the number in front of $x$ .
Doing half of the negative reciprocal. From a gradient of 2 the answers $-2$ (sign changed, not inverted) and $\frac{1}{2}$ (inverted, sign left alone) both appear on scripts; the perpendicular gradient is $-\frac{1}{2}$ .

Write $m_1 m_2 = -1$ on the page and solve it rather than doing it in your head. With $m_1 = 2$ that reads $2m_2 = -1$ , so $m_2 = -\frac{1}{2}$ , and the sign takes care of itself.
Losing the sign when the original gradient is already negative, so the line perpendicular to $y = -4x + 1$ is given gradient $-\frac{1}{4}$ .

Two negatives cancel: $-1 \div -4 = \frac{1}{4}$ . Sanity-check with the picture, since one of a perpendicular pair goes uphill and the other goes downhill, so the two gradients can never share a sign.
Finding the perpendicular gradient correctly and then keeping the original line's intercept, answering $y = -\frac{1}{2}x + 7$ for the line perpendicular to $y = 2x + 7$ through $(4, 1)$ .

The intercept has to come from the point you were given. Substitute $x = 4$ and $y = 1$ into $y = -\frac{1}{2}x + c$ to get $c = 3$ . Only the gradient is inherited from the first line, never $c$ .
On a perpendicular bisector, substituting one of the two endpoints instead of the midpoint, or finding the midpoint and then using the gradient of the segment itself.

Do the two halves separately and label them: midpoint of $AB$ first, gradient of $AB$ second, then negative reciprocal, then substitute the midpoint. The bisector passes through the midpoint and through neither $A$ nor $B$ .

Frequently asked questions

Are perpendicular lines on the Foundation paper?: No. Spec point A9 asks both tiers to use $y = mx + c$ to identify parallel lines, but the perpendicular part of that spec point is Higher only, and so is the perpendicular bisector work that goes with it. Foundation candidates still meet parallel lines and finding the equation of a line through two given points.
Do I get the perpendicular gradient rule on the formulae sheet?: No. The Exam Aid sheet issued with each 1MA1 paper carries the area of a trapezium, volume of a prism, the circle formulae, the quadratic formula, Pythagoras' theorem and the trigonometric ratios, the sine and cosine rules, the area of a triangle, compound interest and two probability rules. Nothing on it concerns gradients, so $m_1 m_2 = -1$ and the gradient formula both have to be recalled.
How can I tell whether two lines are perpendicular from their equations?: Rearrange both into $y = mx + c$ and multiply the two gradients. A product of $-1$ means perpendicular, and equal gradients mean parallel. For example $3x + y = 5$ has gradient $-3$ and $x - 3y = 6$ has gradient $\frac{1}{3}$ , and $-3 \times \frac{1}{3} = -1$ , so those two are perpendicular.
What is the perpendicular bisector of a line segment?: It is the line that crosses the segment at right angles through its midpoint, so every point on it is the same distance from both ends. To find its equation, work out the midpoint, work out the gradient of the segment, take the negative reciprocal for the bisector's gradient, then substitute the midpoint to find $c$ .

Parallel and Perpendicular Lines - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics