Lesson 19 of 20

Parallel and Perpendicular Lines

Discover the elegant relationships between gradients. Parallel lines share the same slope; perpendicular lines have slopes that multiply to negative one.

45 min 5 MCQ + 3 SAQ MAS-LIN-C-02
parallel: m = 2 parallel: m = 2 m = 0 (horizontal) undefined (vertical) perpendicular
Think First

Before we begin: Line $A$ has gradient $2$. Line $B$ is parallel to $A$. Line $C$ is perpendicular to $A$. What can you say about the gradients of $B$ and $C$?

Come back to this after you have worked through the lesson.

1
The Big Idea

Parallel and perpendicular lines are everywhere in geometry and design. Parallel lines never meet because they share the same gradient. Perpendicular lines meet at right angles because their gradients have a special multiplicative relationship. Once you know these properties, you can find the equation of any line parallel or perpendicular to a given line through any point.

Know
Parallel lines have equal gradients. Perpendicular lines have gradients that are negative reciprocals.
Understand
Why perpendicular gradients multiply to $-1$. How to use a given line to find a parallel or perpendicular line through a point.
Can Do
Find the equation of a parallel or perpendicular line through a given point. Classify the relationship between two lines.
2
Lesson Objectives
  • State that parallel lines have the same gradient.
  • Find the perpendicular gradient as the negative reciprocal.
  • Write the equation of a parallel or perpendicular line through a given point.
  • Classify the relationship between two given lines as parallel, perpendicular, or neither.
3
Key Vocabulary
Parallel lines
Lines with the same gradient that never meet ($m_1 = m_2$).
Perpendicular lines
Lines that meet at right angles ($m_1 \times m_2 = -1$).
Negative reciprocal
The perpendicular gradient is $-\dfrac{1}{m}$ (flip the fraction and change the sign).
4
Spot the Trap

Wrong: Thinking perpendicular gradients are just negatives. If $m = 2$, the perpendicular gradient is not $-2$.

Right: Perpendicular gradient is $-\dfrac{1}{2}$ (flip to $\dfrac{1}{2}$, then change sign).

Wrong: Forgetting to take the negative of the reciprocal. $m = -\dfrac{2}{3}$ gives perpendicular $m = \dfrac{2}{3}$ (wrong).

Right: Perpendicular to $m = -\dfrac{2}{3}$ is $m = \dfrac{3}{2}$. Flip to $-\dfrac{3}{2}$, then negate to $\dfrac{3}{2}$.

5
Parallel Lines
+5 XP

Parallel lines never meet. They point in the same direction, so they share the same gradient. If two lines are parallel, their gradients are equal. To find the equation of a line parallel to a given line through a point: copy the gradient, substitute the point, and solve for $c$.

$m_1 = m_2$. Same direction. Never meet.

m = 3 m = 3 m_1 = m_2
copy the gradient
Same gradient rule
Parallel lines have $m_1 = m_2$. The y-intercepts can be different.
Find the equation
Use $y = mx + c$ with the copied gradient. Substitute the point to find $c$.
Horizontal and vertical
Horizontal lines ($m = 0$) are parallel to each other. Vertical lines are parallel to each other.
6
Perpendicular Lines
+5 XP

Perpendicular lines meet at right angles. Their gradients have a special multiplicative relationship: the product of their gradients equals $-1$. Equivalently, the perpendicular gradient is the negative reciprocal of the original. Flip the fraction and change the sign.

$m_1 \times m_2 = -1$. Flip and negate. $m_2 = -\dfrac{1}{m_1}$.

m = 2 m = -1/2 2 x (-1/2) = -1
flip the fraction, change sign
Product test
Multiply the two gradients. If the product is $-1$, the lines are perpendicular.
Negative reciprocal
$m = 2 \rightarrow -\dfrac{1}{2}$. $m = -\dfrac{3}{4} \rightarrow \dfrac{4}{3}$. $m = -1 \rightarrow 1$.
Horizontal and vertical
A horizontal line ($m = 0$) and a vertical line are perpendicular. Their gradients do not multiply to $-1$ because vertical lines have undefined gradient.
7
Classifying Relationships
+5 XP

Given two lines, you can determine if they are parallel, perpendicular, or neither by comparing their gradients. First convert both equations to gradient-intercept form $y = mx + c$ if necessary, then extract the gradients and apply the tests.

Parallel: $m_1 = m_2$. Perpendicular: $m_1 \times m_2 = -1$. Neither: neither test passes.

Find m_1 and m_2 m_1 = m_2? Parallel m_1 x m_2 = -1? Perpendicular Neither: different
compare gradients
Convert first
If lines are in general form, rearrange to $y = mx + c$ to read the gradients clearly.
Test parallel first
Check $m_1 = m_2$ before checking the product. If they are equal, they cannot be perpendicular.
Fractions need care
Simplify fractions before comparing. $\dfrac{4}{10} = \dfrac{2}{5}$, so lines with these gradients are parallel.
8
Putting It All Together
+5 XP

Many geometry problems require you to combine parallel and perpendicular properties. Rectangles have opposite sides parallel and adjacent sides perpendicular. Squares and rhombuses have similar relationships. When asked to find equations in a geometric figure, identify which sides are parallel or perpendicular, then apply the gradient rules.

Opposite sides parallel. Adjacent sides perpendicular. Use gradient rules.

AB || CD AB || CD AB perp BC AB perp BC m_AB = m_CD
geometric properties
Identify the shape
Know which sides are parallel and which are perpendicular for common shapes like rectangles and squares.
Missing vertices
For a rectangle with three vertices known, use parallel and perpendicular properties to find the fourth.
Verify with distance
Check that opposite sides have equal length to confirm your rectangle is correct.
Worked Example 1 — Parallel Line
Q.Find the equation of the line parallel to $y = 3x + 5$ passing through $(2, -1)$.
Solution
1

Gradient: Parallel lines share the same gradient, so $m = 3$.

2

Write: $y = 3x + c$

3

Substitute $(2, -1)$: $-1 = 3(2) + c$ → $-1 = 6 + c$ → $c = -7$

4

Answer: $y = 3x - 7$

Answer: $y = 3x - 7$
Worked Example 2 — Perpendicular Line
Q.Find the equation of the line perpendicular to $y = 2x + 1$ passing through $(4, 3)$.
Solution
1

Perpendicular gradient: $m_1 = 2$, so $m_2 = -\dfrac{1}{2}$

2

Write: $y = -\frac{1}{2}x + c$

3

Substitute $(4, 3)$: $3 = -\frac{1}{2}(4) + c$ → $3 = -2 + c$ → $c = 5$

4

Answer: $y = -\frac{1}{2}x + 5$

Answer: $y = -\frac{1}{2}x + 5$
Worked Example 3 — Classify Two Lines
Q.Determine if $2x + 3y = 6$ and $3x - 2y = 4$ are parallel, perpendicular, or neither.
Solution
1

Convert to gradient-intercept form:

2

$2x + 3y = 6$ → $3y = -2x + 6$ → $y = -\frac{2}{3}x + 2$, so $m_1 = -\frac{2}{3}$

3

$3x - 2y = 4$ → $-2y = -3x + 4$ → $y = \frac{3}{2}x - 2$, so $m_2 = \frac{3}{2}$

4

Test: $m_1 \times m_2 = -\frac{2}{3} \times \frac{3}{2} = -1$. The lines are perpendicular.

Answer: The lines are perpendicular.

Brain Trainer

4 quick-fire drills. Beat the clock.

What is the perpendicular gradient to $m = 4$?
$-\dfrac{1}{4}$
Find the equation of the line parallel to $y = 2x + 3$ through $(1, 5)$.
$y = 2x + 3$
Classify $y = 3x + 1$ and $y = -\frac{1}{3}x + 2$.
Perpendicular
Find the equation of the line perpendicular to $y = -\frac{2}{3}x + 1$ through $(3, 0)$.
$y = \frac{3}{2}x - \frac{9}{2}$
Practice

5 MCQs and 3 short-answer questions. Target: 80% accuracy.

1.A line parallel to $y = 4x - 3$ has gradient:
Correct! Parallel lines have equal gradients. The gradient of $y = 4x - 3$ is $4$.
Parallel lines share the same gradient. Read the coefficient of $x$ in the given equation.
2.The gradient of a line perpendicular to $y = -\dfrac{2}{5}x + 1$ is:
Correct! Flip $-\frac{2}{5}$ to $-\frac{5}{2}$, then negate to get $\frac{5}{2}$.
The perpendicular gradient is the negative reciprocal. Flip the fraction, then change the sign.
3.The equation of the line through $(2, 5)$ parallel to $y = 3x + 1$ is:
Correct! $m = 3$, so $5 = 3(2) + c$ → $5 = 6 + c$ → $c = -1$. Thus $y = 3x - 1$.
Copy the gradient ($m = 3$), then substitute the point $(2, 5)$ into $y = 3x + c$ and solve for $c$.
4.Which pair of lines are perpendicular?
Correct! $3 \times (-\frac{1}{3}) = -1$, so these lines are perpendicular.
Multiply the gradients of each pair. Look for a product of $-1$.
5.The equation of the line perpendicular to $y = \dfrac{3}{4}x - 2$ through $(4, 1)$ is:
Correct! Perpendicular gradient is $-\frac{4}{3}$. Then $1 = -\frac{4}{3}(4) + c$ → $1 = -\frac{16}{3} + c$ → $c = \frac{19}{3}$.
First find the perpendicular gradient by flipping and negating. Then substitute the point to find $c$.
6.Find the equation of the line parallel to $y = -2x + 5$ passing through $(-1, 4)$. Give your answer in the form $y = mx + c$.

Your answer:

Parallel gradient identified correctly
Point substituted correctly
Final equation correct
Model Answer

Parallel gradient: $m = -2$

$y = -2x + c$

$4 = -2(-1) + c$ → $4 = 2 + c$ → $c = 2$

$y = -2x + 2$

7.Determine whether each pair of lines is parallel, perpendicular, or neither. Show your reasoning.

(a) $y = 3x + 2$ and $y = -\dfrac{1}{3}x - 1$

(b) $2x - 5y = 10$ and $4x - 10y = 20$

(c) $x + 2y = 6$ and $2x - y = 3$

Your answer:

Each pair classified correctly
Gradient calculations shown
Reasoning clear for each case
Model Answer

(a) $m_1 = 3$, $m_2 = -\frac{1}{3}$. $3 \times (-\frac{1}{3}) = -1$. Perpendicular.

(b) $2x - 5y = 10$ → $y = \frac{2}{5}x - 2$. $4x - 10y = 20$ → $y = \frac{4}{10}x - 2 = \frac{2}{5}x - 2$. Same gradient. Parallel.

(c) $x + 2y = 6$ → $y = -\frac{1}{2}x + 3$. $2x - y = 3$ → $y = 2x - 3$. $(-\frac{1}{2}) \times 2 = -1$. Perpendicular.

8.A rectangle has vertices $A(1, 2)$, $B(5, 2)$, and $C(5, 6)$.

(a) Find the equation of side $AB$.

(b) Find the equation of the line through $D$ (the fourth vertex) that is perpendicular to $AB$.

(c) Explain why adjacent sides of a rectangle must be perpendicular, and verify this for two adjacent sides of this rectangle.

Your answer:

Equation of AB correct (1 mark)
Equation through D with reasoning (2 marks)
Geometric explanation with verification (2 marks)
Model Answer

(a) $AB$ is horizontal: $y = 2$

(b) $D = (1, 6)$. Perpendicular to $AB$ (horizontal) is vertical: $x = 1$.

(c) Adjacent sides of a rectangle meet at right angles, so they must be perpendicular. $AB$ is horizontal ($m = 0$) and $BC$ is vertical (undefined gradient). A horizontal line and a vertical line are perpendicular because they meet at $90°$.

Review

Consolidate and reflect before moving on.

Stretch Challenge

Find the equation of the line through $(2, -3)$ that is perpendicular to $3x + 4y = 12$. Give your answer in general form. (Hint: first find the gradient of $3x + 4y = 12$ by rearranging to $y = mx + c$.)

Key Idea

Parallel lines have equal gradients ($m_1 = m_2$). Perpendicular lines have gradients that multiply to $-1$ ($m_2 = -\frac{1}{m_1}$). To classify two lines, convert to $y = mx + c$ and compare.

Common Trap

Thinking perpendicular means just negative. If $m = 2$, the perpendicular is $-\frac{1}{2}$, not $-2$. Always flip the fraction first, then change the sign.

Connection

This lesson builds on equations of lines by adding geometric relationships. The final lesson brings everything together with modelling and problem solving.

Interactive: Parallel and Perpendicular Explorer — drag a line and watch its parallel and perpendicular companions update in real time.
Parallel Pro
Perpendicular Master
Shape Classifier
Daily Challenge

Given $y = \frac{2}{3}x - 4$, find (a) the equation of the parallel line through $(3, 2)$, and (b) the equation of the perpendicular line through $(3, 2)$. Time yourself — can you do both in under 2 minutes?