Explore perpendicular lines | Mathematics

Unit 1 · Straight-line graphs

1 Explore Perpendicular Lines

Two roads cross. Sometimes they cross at a perfect corner — a right angle — and sometimes they just slant into each other. Today you learn a secret number that tells you, without a protractor, whether two lines make a perfect corner. The number is −1.

🎯 By the end of this you can…

✓Say what perpendicular means, and spot perpendicular lines on a grid.
✓Discover that perpendicular gradients multiply to give −1.
✓Find the negative reciprocal of any number — whole, fraction or decimal.
✓Write the equation of a line perpendicular to a given line.

💡 The idea

What does perpendicular mean?

Two lines are perpendicular when they meet at a right angle — exactly 90°, a perfect corner, like the corner of this page. On a drawing we mark it with a little square in the corner:

A vertical line and a horizontal line are always perpendicular. But most perpendicular pairs are slanted — and for those, your eyes can be fooled. We need a calculation.

✅ Tick the perpendicular pairs

Tap every pair of lines that meets at a right angle. Then press Check.

💡 The idea

First, remember gradient

The gradient of a line is how steep it is: $gradient = \frac{rise}{run}$ — how far up for every 1 across. Up 4 for every 1 across is gradient 4. Down 1 for every 4 across is gradient $- \frac{1}{4}$ — down means negative.

Now the experiment. Below, line A has a gradient you choose. Line B is drawn at a right angle to A. Look at the two gradients — and at what happens when you multiply them.

💡 The idea

Find a gradient yourself — click two points

Before the experiment, make sure the formula is in your fingers. Tap any two grid corners below: the run appears, then the rise, then the formula fills itself in with your numbers. Try an uphill pair, a downhill pair, and a pair where you go right-to-left.

📐 Gradient finder

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Tap a grid corner: that is point A.

Gradient is rise over run: how far up the line goes for how far across. Pick any two points on a line and the formula does the rest.

🔬 Gradient lab

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Line A goes through the origin. Line B is drawn at a right angle to it — see the little square where they cross. Change the gradient of A and watch B.

gradient of A

gradient of B

−¼

A × B

4 × (−¼) = −1

📋 Is there a connection? Yes — always

1
Line A had gradient 4. The line at right angles to it had gradient $- \frac{1}{4}$ .
2
Multiply them: $4 \times (- \frac{1}{4}) = - 1$ .
3
Try 2 and $- \frac{1}{2}$ : $2 \times (- \frac{1}{2}) = - 1$ . Try 1 and −1: $1 \times (- 1) = - 1$ .
4
The rule: if two lines are perpendicular, the product of their gradients is −1. And it works backwards too: if the product is −1, the lines are perpendicular.

m_{1} \times m_{2} = - 1 ⟺ the lines are perpendicular

✏️ Worked example

Two lines P and Q are drawn. P goes up 1 for every 1 across; Q goes down 1 for every 1 across. Are they perpendicular?

1
Gradient of P: up 1, across 1 → $m_{P} = 1$ .
2
Gradient of Q: down 1, across 1 → $m_{Q} = - 1$ .
3
Multiply: $1 \times (- 1) = - 1$ .
4
The product is −1, so yes — P and Q are perpendicular. (You can also see the little right-angle square where they cross.)

✅ Answer: Yes: m_P × m_Q = 1 × (−1) = −1

💡 The idea

Read the gradients, then multiply — worksheet question 3

Each grid shows a pair of perpendicular lines. Count rise over run for each line (pick two points where the line crosses grid corners), then multiply the two gradients.

🎮 Question 3 · gradients of perpendicular pairs

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a) gradient of L₁ (m₁)

a) gradient of L₂ (m₂)

a) m₁ × m₂

b) gradient of L₃ (m₃)

b) gradient of L₄ (m₄)

b) m₃ × m₄

c) gradient of L₅ (m₅)

c) gradient of L₆ (m₆)

c) m₅ × m₆

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⚠️ The classic trap

Rise is a signed number. A line going down as you move right has a negative gradient. If you forget the minus, your product comes out as +1 and you will wrongly say the lines are not perpendicular.

Say it out loud as you read a grid: “down 1, across 3 — minus one third.”

💡 The idea

Have a think: what do you multiply by to get 1?

For each number, what do you need to multiply it by to get 1? $3 \times □ = 1$ → the answer is $\frac{1}{3}$ , because $3 \times \frac{1}{3} = 1$ .

$\frac{3}{4} \times □ = 1$ → flip it: $\frac{4}{3}$ , because $\frac{3}{4} \times \frac{4}{3} = \frac{12}{12} = 1$ .

This flipped number has a name: the reciprocal. When you multiply a number by its reciprocal you always get 1. A whole number like 3 is $\frac{3}{1}$ , so its reciprocal is $\frac{1}{3}$ .

🎮 Reciprocals · multiply to get 1

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4 \times □ = 1

\frac{2}{3} \times □ = 1

\frac{1}{5} \times □ = 1

0.5 \times □ = 1

7 \times □ = 1

0.25 \times □ = 1

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📋 Negative reciprocal — the same idea, with a minus sign

1
To get 1 from 3 you multiply by $\frac{1}{3}$ . That is the reciprocal.
2
To get −1 from 3 you multiply by $- \frac{1}{3}$ . That is the negative reciprocal: flip it, and change the sign.
3
The negative reciprocal of $\frac{3}{4}$ is $- \frac{4}{3}$ . The negative reciprocal of $- \frac{2}{5}$ is $+ \frac{5}{2}$ — the minus sign flips too!
4
Why we care: perpendicular gradients multiply to −1. So the gradient of a perpendicular line is the negative reciprocal of the first gradient. Flip and change sign — that is the whole skill.

🎮 Question 4 · complete the calculations (□ × ? = −1)

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□ \times 6 = - 1

3 \times □ = - 1

- 2 \times □ = - 1

- \frac{1}{7} \times □ = - 1

\frac{2}{3} \times □ = - 1

- 1 = \frac{5}{7} \times □

- 1 = - 9 \times □

- 0.5 \times □ = - 1

□ \times 0.25 = - 1

□ \times (- 0.625) = - 1

- a \times □ = - 1

\frac{3 b}{5} \times □ = - 1

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🎮 Question 7 · write the negative reciprocal

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\frac{5}{3}

- \frac{2}{5}

\frac{4}{7}

13

- \frac{1}{14}

- 0.8

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🗂️ Perpendicular or not? Sort the gradient pairs

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Tap a card, then tap the box you think it belongs in.

🤔 Tommy says: “When two lines are perpendicular, one will always have a positive gradient and the other will always have a negative gradient.” Do you agree?

💡 The idea

Before perpendicular equations: y = mx + c

Every straight line has an equation $y = m x + c$ . Two numbers say everything: m, the gradient (the number multiplying $x$ ), and c, where the line cuts the y-axis. Play with both below until you can predict the picture before it appears.

🧪 Line lab · y = mx + c

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y = 2x + 1

m — the gradient (the number with x)

c — the y-intercept (where it cuts the y-axis)

• The ring sits at (0, 1): change c and the whole line slides up or down — it never tilts.

• The little triangle: across 1, up 2. Change m and the line tilts about the ring — it never slides.

🎮 Read m and c straight off an equation

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m for

y = 3 x - 2

c for

y = 3 x - 2

m for

y = 5 - 2 x

c for

y = 5 - 2 x

m for

2 y = 4 x + 6

c for

2 y = 4 x + 6

m for

y = - x + 4

c for

y = \frac{1}{2} x

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💡 The idea

Find the equation from two points — the full workout

A line through two points has one equation, and you can find it in two steps: step 1, the gradient from rise over run (you did this in the Gradient finder); step 2, put one of the points into $y = m x + c$ and solve for c. Tap two corners and watch both steps happen with your own numbers.

🧭 Equation from two points

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Tap a grid corner: point A.

Two points fix a line. From them: m is rise over run; then put one point into $y = m x + c$ to find c.

✏️ Worked example

Find the equation of the line through (1, 5) and (3, 11).

1
Step 1 · gradient. run = 3 − 1 = 2, rise = 11 − 5 = 6, so $m = \frac{6}{2} = 3$ .
2
Step 2 · c. The line is $y = 3 x + c$ . Put in (1, 5): $5 = 3 \times 1 + c$ , so $c = 2$ .
3
Check with the other point: x = 3 gives 3 × 3 + 2 = 11 ✓.

✅ Answer: $y = 3 x + 2$

🎮 Equation from two points

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m for the line through (0, 2) and (4, 10)

c for the line through (0, 2) and (4, 10)

m for the line through (2, 1) and (4, −3)

c for the line through (2, 1) and (4, −3)

equation of the line through (2, 1) and (4, −3) (write as y = …)

c for the line through (−1, 4) and (1, 0)

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📋 Writing the equation of a perpendicular line

1
Start with $y = 2 x + 5$ . Its gradient is the number in front of $x$ : 2.
2
A perpendicular line has the negative reciprocal gradient: $- \frac{1}{2}$ .
3
So any line $y = - \frac{1}{2} x + c$ is perpendicular to it — $y = - \frac{1}{2} x + 1$ , $y = - \frac{1}{2} x - 7$ , whichever you like. There are infinitely many, all parallel to each other.
4
What if it had been $y = 3 - 2 x$ ? Careful — the gradient is −2 (the number stuck to $x$ , minus sign included). Negative reciprocal of −2 is $+ \frac{1}{2}$ : $y = \frac{1}{2} x + c$ .

💡 The idea

Question 5 · the line y = 4x — watch M and N appear

The red line is $y = 4 x$ : gradient 4, through the origin.

A line perpendicular to it has gradient $- \frac{1}{4}$ (the negative reciprocal of 4).

✏️ Worked example

Question 9 · Write the equation of the line perpendicular to $y = 8 - 3 x$ that passes through (0, 15).

1
Gradient of the given line: the number with $x$ is −3.
2
Perpendicular gradient = negative reciprocal of −3 = $+ \frac{1}{3}$ .
3
“Passes through (0, 15)” means it crosses the y-axis at 15 — so $c = 15$ .
4
Put them together. This time there is only one answer, because the point pins down $c$ .

✅ Answer: $y = \frac{1}{3} x + 15$

✏️ Worked example

Question 10 · $L_{1} : y - 6 x = 15$ and $L_{2} : 6 y = 3 - x$ . Show that L₁ and L₂ are perpendicular.

1
Get each into the form $y = m x + c$ so the gradient can be read.
2
L₁: add $6 x$ to both sides → $y = 6 x + 15$ . Gradient $m_{1} = 6$ .
3
L₂: divide everything by 6 → $y = \frac{3}{6} - \frac{1}{6} x = - \frac{1}{6} x + \frac{1}{2}$ . Gradient $m_{2} = - \frac{1}{6}$ .
4
Multiply: $6 \times (- \frac{1}{6}) = - 1$ . The product is −1, so the lines are perpendicular. ∎

✅ Answer: m₁ × m₂ = 6 × (−⅙) = −1, so L₁ ⟂ L₂

⚠️ The classic trap

Rearrange first, read second. In $6 y = 3 - x$ the gradient is not −1 — you must divide by 6 first. And in $y = 8 - 3 x$ the gradient is −3, not 8: the gradient is whatever is multiplying $x$ , with its sign.

🎮 Questions 5, 8, 9 · gradients and equations

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Q5a) gradient of

y = 4 x

Q5b) gradient of a line perpendicular to

y = 4 x

Q5c) equation of M: perpendicular to

y = 4 x

, through the origin (write as y = …)

Q5d) equation of N: perpendicular to

y = 4 x

, through (0, −2)

Q5e) Are M and N perpendicular to each other? (yes/no)

Q8) gradient of any line perpendicular to

y = 14 x + 1

Q9) gradient of the line perpendicular to

y = 8 - 3 x

Q9) equation of that line through (0, 15) (write as y = …)

Q10) gradient of

6 y = 3 - x

Q10) m₁ × m₂ for

y - 6 x = 15

and

6 y = 3 - x

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🏅

You finished this lesson!

Perpendicular lines meet at 90°. Their gradients multiply to −1, so a perpendicular gradient is the negative reciprocal: flip it, change the sign. Read the gradient carefully (rearrange to

y = m x + c

first), and remember that lines with the same gradient are parallel, not perpendicular. Now try the quiz — every question has a notebook page for your working.

Keep reading — this lesson marks itself done automatically.