Perpendicular Lines

Perpendicular lines have gradients that multiply to −1: flip and change the sign, then find the line's equation or a missing coordinate.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1, Edexcel Further Pure Maths 4PM1 · 53 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Flip and Change the Sign

Perpendicular lines meet at a right angle. Their gradients are linked: multiply them and you always get −1.

Formula

m_{1} \times m_{2} = - 1

So to get the perpendicular gradient, turn the fraction upside down and change its sign.

Gradient	Perpendicular gradient
$2$	$- \frac{1}{2}$
$\frac{3}{4}$	$- \frac{4}{3}$

Rule

Check: $2 \times (- \frac{1}{2}) = - 1$ . If the product is not −1, the lines are not perpendicular.

Equation of a Perpendicular Line

The 4-mark question: “Find an equation of the line perpendicular to L through a point.” Often it then asks where that line crosses the x-axis.

Worked example

1
$L : y = 2 x + 1$ — L is in the form y = mx + c, so its gradient is the number in front of x: 2.
2
$m = - \frac{1}{2}$ — Write 2 as $\frac{2}{1}$ , flip it and change the sign. Check: $2 \times (- \frac{1}{2}) = - 1$ .
3
$y = - \frac{1}{2} x + c$ — The new line has this gradient. Its c is still unknown, so leave it as c for now.
4
$3 = - \frac{1}{2} \times 4 + c$ — The line passes through (4, 3), so put x = 4 and y = 3 in.
5
$3 = - 2 + c$ — $- \frac{1}{2} \times 4 = - 2$ . Keep the minus sign.
6
$c = 5$ — Add 2 to both sides: 3 + 2 = 5.
7
$y = - \frac{1}{2} x + 5$ — Put c back in. This is the answer in the form y = mx + c.
8
$2 y = - x + 10$ — Only if the question wants no fractions: multiply every term by 2.
9
$x + 2 y - 10 = 0$ — Add x and take 10 from both sides, so every term is on one side.

Key idea

Crossing the x-axis means $y = 0$ : $0 = - \frac{1}{2} x + 5$ , so $\frac{1}{2} x = 5$ and $x = 10$ . The point is $(10, 0)$ .

Common mistake

Using L's gradient 2 for the new line. That gives a parallel line, not a perpendicular one.

Starting From Two Points

Most exam questions give the first line as two points, A and B. Find the gradient of AB first, then carry on as before.

Find the line perpendicular to AB through (4, 1), where A is (−1, 6) and B is (3, −2).

Worked example

1
$m_{A B} = \frac{- 2 - 6}{3 - ( - 1 )}$ — Gradient = change in y ÷ change in x. Take B's numbers minus A's, in the same order on top and bottom. Put −1 in brackets.
2
$= \frac{- 8}{4}$ — −2 − 6 = −8. And 3 − (−1) = 3 + 1 = 4, because taking away a negative adds.
3
$m_{A B} = - 2$ — −8 ÷ 4 = −2. AB goes down from left to right.
4
$m = \frac{1}{2}$ — Flip −2 to $- \frac{1}{2}$ , then change the sign to $+ \frac{1}{2}$ . Check: $- 2 \times \frac{1}{2} = - 1$ .
5
$1 = \frac{1}{2} \times 4 + c$ — The line passes through (4, 1), so put x = 4 and y = 1 into $y = \frac{1}{2} x + c$ .
6
$1 = 2 + c$ — $\frac{1}{2} \times 4 = 2$ .
7
$c = - 1$ — Take 2 from both sides: 1 − 2 = −1.
8
$y = \frac{1}{2} x - 1$ — Put c back in. Check with the point: $\frac{1}{2} \times 4 - 1 = 1$ . It works.

Common mistake

Writing 3 − 1 = 2 on the bottom. It is 3 − (−1), which is 3 + 1 = 4. Brackets round negatives stop this.

Key idea

Show every line. Most of the marks are for the working: the gradient of AB, the flipped gradient, and putting the point in. A slip in the last line can still leave you with 3 marks out of 4.

When y Is Not on Its Own

The 2-mark question: “Find the gradient of a line perpendicular to $3 y + 4 x = 6$ .” First get y alone to read the gradient.

Worked example

1
$3 y + 4 x = 6$ — The question. y has a 3 in front and an x term beside it, so the gradient is not on show yet.
2
$3 y = - 4 x + 6$ — Take 4x from both sides, so only the y term is left on the left.
3
$y = - \frac{4}{3} x + 2$ — Divide every term by 3: $- 4 x \div 3 = - \frac{4}{3} x$ and $6 \div 3 = 2$ . Now the number in front of x is the gradient.
4
$m = \frac{3}{4}$ — Flip and change the sign: $\frac{4}{3}$ becomes $\frac{3}{4}$ , and minus becomes plus.
5
$- \frac{4}{3} \times \frac{3}{4} = - 1$ — Check: the product is −1, so the lines are perpendicular.

Common mistake

Reading the gradient as 4, the number in front of x. Until y is alone, that number is not the gradient.

Key idea

For more on rearranging, see The Equation y = mx + c.

Finding a Missing Coordinate

The 4-mark question: “L is perpendicular to M. Work out the value of k.” Write M's gradient in two ways and make them equal.

Worked example

1
$L : y = 3 x - 1$ — L is in the form y = mx + c, so its gradient is 3.
2
$M gradient = - \frac{1}{3}$ — M is perpendicular to L: flip 3 to $\frac{1}{3}$ and change the sign.
3
$A (2, 5), B (8, k) on M$ — Both points lie on M, so the gradient from A to B must also be $- \frac{1}{3}$ .
4
$\frac{k - 5}{8 - 2} = - \frac{1}{3}$ — Gradient = change in y ÷ change in x, written with k. Set it equal to M's gradient.
5
$\frac{k - 5}{6} = - \frac{1}{3}$ — Work out the bottom: 8 − 2 = 6.
6
$k - 5 = - 2$ — Multiply both sides by 6: $- \frac{1}{3} \times 6 = - 2$ .
7
$k = 3$ — Add 5 to both sides: −2 + 5 = 3.
8
$\frac{3 - 5}{6} = - \frac{2}{6} = - \frac{1}{3}$ — Check: put k = 3 back. The gradient is $- \frac{1}{3}$ , as it should be.

Key idea

A right angle in a triangle, like angle ABC = 90°, works the same way: the gradients of AB and BC multiply to −1.

Tangent to a Circle

A tangent touches a circle at one point, P. It is always perpendicular to the radius at P. So the radius gives you the gradient.

A circle has centre C(1, 2). P(7, 6) is on the circle. Find the equation of the tangent at P.

Worked example

1
$m_{C P} = \frac{6 - 2}{7 - 1}$ — The radius joins the centre C to P. Gradient = change in y ÷ change in x.
2
$= \frac{4}{6} = \frac{2}{3}$ — 6 − 2 = 4 and 7 − 1 = 6. Divide top and bottom by 2: the radius gradient is $\frac{2}{3}$ .
3
$m = - \frac{3}{2}$ — The tangent is perpendicular to the radius: flip and change the sign. Check: $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ .
4
$6 = - \frac{3}{2} \times 7 + c$ — The tangent passes through P(7, 6), so put x = 7 and y = 6 in.
5
$6 = - \frac{21}{2} + c$ — $- \frac{3}{2} \times 7 = - \frac{21}{2}$ , which is −10.5.
6
$c = \frac{33}{2}$ — Add $\frac{21}{2}$ to both sides: 6 + 10.5 = 16.5, which is $\frac{33}{2}$ .
7
$y = - \frac{3}{2} x + \frac{33}{2}$ — Put c back in. This is the tangent at P.
8
$2 y = - 3 x + 33$ — Only if the question wants no fractions: multiply every term by 2. Both forms are correct.

Key idea

Sometimes you are given A, the other end of the diameter, not P. The centre is halfway, so repeat the step. From A(−5, −2) to C(1, 2) is 6 right and 4 up, so P is (1 + 6, 2 + 4) = (7, 6).

Common mistake

Using the radius gradient $\frac{2}{3}$ for the tangent. The tangent is at 90° to the radius, so flip and change the sign.

Which Method? Read the Question

Find the gradient you are given first. Then decide what to do with it.

You see	You do
Two points, A and B	gradient of AB first: change in y ÷ change in x
“Perpendicular gradient”	get y alone, flip, change sign
“Perpendicular … passes through”	new gradient, then find c
Through the origin	c = 0
“Crosses the x-axis”	put y = 0
An unknown like k	gradient two ways, make them equal
Tangent to a circle at P	radius gradient, flip, change sign
Perpendicular to a flat line like y = 3	an upright line: x = the point's x
“Parallel”	Parallel Lines and Their Equations
Through the midpoint of AB	The Perpendicular Bisector

Key idea

Always check: the two gradients multiplied must give exactly −1.

Practise perpendicular lines now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on perpendicular lines

Each with a full worked solution.

All 53 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.