How do you check whether two lines are perpendicular?

For two non-vertical lines, multiply their gradients. If the product is $-1$, they are perpendicular. A horizontal line and a vertical line are also perpendicular, but the vertical line has an undefined gradient.

What does negative reciprocal mean?

It means taking the reciprocal and changing the sign. For example, the negative reciprocal of $\frac{3}{4}$ is $-\frac{4}{3}$. Check that the original gradient and your answer multiply to $-1$.

Do perpendicular lines have the same intercept?

No. Perpendicularity determines the new gradient, not the intercept. Substitute the point given in the question into the new equation to calculate its intercept.

Are perpendicular line equations foundation or higher tier?

Using perpendicular gradients to identify lines and find equations is higher-tier GCSE content. Recognising perpendicular lines as lines meeting at a right angle is taught earlier.

Common Mistakes With Perpendicular Lines in GCSE Maths

A question can look familiar and still cost marks. You recognise a straight-line equation, remember something about flipping the gradient, then hesitate over the minus sign. The common mistakes with perpendicular lines in GCSE Maths usually come down to three things: reading the original gradient incorrectly, taking the negative reciprocal incorrectly, or stopping before finding the complete equation.

The repair is a repeatable method: rearrange, identify the gradient, take its negative reciprocal, then use the given point. Questions using perpendicular gradients belong to GCSE higher tier, although recognising perpendicular lines starts much earlier.

Imagine a student checking a practice paper. Rather than writing “revise graphs” beside every wrong answer, they identify the exact step that failed. That is the approach we will follow here.

Your perpendicular lines checklist

Perpendicular lines meet at a right angle of $90∘90^\circ$ . For two non-vertical perpendicular lines, their gradients satisfy:

m_1m_2=-1

In other words, one gradient is the negative reciprocal of the other.

Before answering, check that you can:

Rearrange a straight-line equation into $y = m x + c$ .
Identify $m$ , the gradient, without confusing it with $c$ , the vertical intercept.
Find the negative reciprocal, changing both the sign and the reciprocal.
Substitute a given point to find the new intercept.
Check the gradients and the completed equation.

The parallel and perpendicular lines revision guide brings these ideas together if you need a reference alongside your practice.

Common mistakes with perpendicular lines in GCSE Maths

Mistake: changing the sign but not taking the reciprocal

Our hypothetical student sees a line with gradient $3$ and writes that its perpendicular gradient is $- 3$ .

Why it happens: “negative reciprocal” gets shortened in memory to “make it negative”. But changing the sign is only half the operation.

Correct method: write the whole number as a fraction, take its reciprocal, then change its sign:

3=31⟹m⊥=−133=\frac{3}{1} \quad\Longrightarrow\quad m_{\perp}=-\frac{1}{3}

Check by multiplying:

3×(−13)=−13\times\left(-\frac{1}{3}\right)=-1

The incorrect choice would give $3×(−3)=−93\times(-3)=-9$ , so it fails the check.

Mistake: taking the reciprocal but keeping the sign

Next, the student meets a gradient of $−25-\frac{2}{5}$ and chooses $−52-\frac{5}{2}$ .

Why it happens: flipping the fraction feels like the main task, so the sign is overlooked.

Correct method: reverse the numerator and denominator, then change the sign:

m=−25⟹m⊥=52m=-\frac{2}{5} \quad\Longrightarrow\quad m_{\perp}=\frac{5}{2}

A negative non-zero gradient has a positive perpendicular gradient. A positive gradient has a negative perpendicular gradient. This sign check is quick enough to use before doing any multiplication.

Mistake: reading the gradient before rearranging

Suppose the question gives:

2 x + 3 y = 12

The student selects $2$ as the gradient because it is beside $x$ .

Why it happens: the rule “the gradient is the coefficient of $x$ ” only works when the equation has been written as $y = m x + c$ .

Correct method: isolate $y$ first:

3 y = - 2 x + 12

y=−23x+4y=-\frac{2}{3}x+4

The original gradient is $−23-\frac{2}{3}$ , so the perpendicular gradient is $32\frac{3}{2}$ .

Notice that every term is divided by $3$ . Dividing only the term containing $x$ would leave an incorrect intercept, even if the gradient happened to be right.

Worked example: finish the whole equation

Finding the gradient is not always the finish line. If the question asks for an equation, a gradient alone does not identify one particular line.

Question: Find the equation of the line perpendicular to $y = 2 x + 5$ that passes through $(4, 1)$ .

Identify the new gradient

The original gradient is $2$ . Its negative reciprocal is:

m⊥=−12m_{\perp}=-\frac{1}{2}

Write the new line with an unknown intercept:

y=−12x+cy=-\frac{1}{2}x+c

Substitute the given point

The point $(4, 1)$ means $x = 4$ and $y = 1$ :

1=−12(4)+c1=-\frac{1}{2}(4)+c

1 = - 2 + c

c = 3

Therefore:

y=−12x+3\boxed{y=-\frac{1}{2}x+3}

Check both conditions

The gradients multiply to $- 1$ :

2×(−12)=−12\times\left(-\frac{1}{2}\right)=-1

The given point also satisfies the new equation:

−12(4)+3=1-\frac{1}{2}(4)+3=1

These checks test different things. The first checks perpendicularity; the second checks that you have chosen the correct line from all the possible perpendicular lines.

Use MathsGenie's parallel and perpendicular lines Lessons and perpendicular lines Question Bank to practise this same sequence without having to assemble your own questions.

Stick-figure comic showing a student sending their work through a sign-check security checkpoint

More perpendicular lines mistakes to catch

Keeping the original intercept

In the worked example, copying $+ 5$ from the original equation would produce the wrong line.

Why it happens: the student changes the gradient and assumes everything else stays the same. But perpendicularity determines the gradient, not the intercept. Use the supplied point to calculate $c$ afresh.

Confusing parallel and perpendicular lines

Parallel lines have equal gradients. Non-vertical perpendicular lines have gradients whose product is $- 1$ .

Before calculating, underline the relationship requested. A correct parallel-line method cannot answer a perpendicular-line question. The parallel and perpendicular lines topic page is useful for practising the distinction.

Trying to divide by zero

A horizontal line such as $y = 4$ has gradient $0$ . A vertical line has an undefined gradient, so do not try to find it by calculating $−10-\frac{1}{0}$ .

Instead, use the geometry: horizontal and vertical lines are perpendicular. The vertical line through $(3, 4)$ is $x = 3$ .

Exam-style practice

Try these without looking at the answers. Show the gradient change explicitly, then show any substitution needed.

Gradient check: Find the gradient of a line perpendicular to $y = - 4 x + 7$ .
Rearranging check: Find the gradient of a line perpendicular to $3 x + 2 y = 8$ .
Complete equation: Find the equation of the line perpendicular to $y=13x−2y=\frac{1}{3}x-2$ through $(2, 5)$ .

Answers and checks

Gradient check: The negative reciprocal of $- 4$ is $14\frac{1}{4}$ .

Rearranging check: Writing $3 x + 2 y = 8$ as $y=−32x+4y=-\frac{3}{2}x+4$ gives a perpendicular gradient of $23\frac{2}{3}$ .

Complete equation: The new gradient is $- 3$ , so write $y = - 3 x + c$ . Substituting $(2, 5)$ gives:

\quad\Longrightarrow\quad c=11

Therefore the equation is $y = - 3 x + 11$ .

If an answer differs, locate the first incorrect step rather than copying the finished solution. Label the error “rearranging”, “negative reciprocal” or “substitution”. That tells you what your next practice question needs to test.

Turn corrections into reliable marks

The student in our example now has something more useful than “be careful”: a specific check for each recurring error. Your next task is to make those checks routine when the question looks unfamiliar. Start your next practice session with MathsGenie's Lessons and Question Bank above, working through perpendicular gradients before moving on to complete equations.