Parallel and Perpendicular Lines

Understanding gradient relationships between lines

SPEC 3.1 • COORDINATE GEOMETRY

Why This Matters

Parallel and perpendicular lines appear everywhere in mathematics and the real world - from architecture and engineering to computer graphics and navigation.

Key Insight

The relationship between two lines can be determined entirely by comparing their gradients. You don't need to graph them!

In This Lesson

Gradient condition for parallel lines

Gradient condition for perpendicular lines

Finding equations of parallel and perpendicular lines

Classifying pairs of lines

Parallel Lines

Parallel lines never meet - they always stay the same distance apart. This happens when they have exactly the same steepness.

Condition

m_{1} = m_{2}

Parallel lines have equal gradients

Example

y = 2 x + 3

and

y = 2 x - 2

are parallel (both have gradient 2)

Strategy

To find a parallel line: keep the same gradient, use point-gradient form with the new point.

Worked Example01

Find the equation of the line parallel to $y = 3 x + 2$ that passes through the point (1, 5).

1

Identify gradient of given line: m = 3

2

Parallel line has same gradient: m = 3

3

Use point-gradient form with (1, 5):

y - 5 = 3 (x - 1)

4

Expand:

y - 5 = 3 x - 3

5

Rearrange:

y = 3 x + 2

Answer: y = 3x + 2

Worked Example02

Find the equation of the line parallel to $2 x + 3 y - 6 = 0$ that passes through (-3, 4). Give your answer in the form $a x + b y + c = 0$ .

1

Find gradient of given line:

3 y = - 2 x + 6

, so

y = - \frac{2}{3} x + 2

, m =

- \frac{2}{3}

2

Parallel line has gradient m =

- \frac{2}{3}

3

Use point-gradient form:

y - 4 = - \frac{2}{3} (x + 3)

4

Multiply by 3:

3 (y - 4) = - 2 (x + 3)

5

Expand:

3 y - 12 = - 2 x - 6

6

Rearrange:

2 x + 3 y - 6 = 0

Answer: 2x + 3y - 6 = 0

Perpendicular Lines

Perpendicular lines meet at right angles (90°). Their gradients have a special relationship.

Condition

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

The gradients multiply to give -1

Negative Reciprocal

If line 1 has gradient

m

, the perpendicular has gradient

- \frac{1}{m}

Examples

m = 2 → perpendicular m = −½

m = −¾ → perpendicular m = ⁴⁄₃

Worked Example03

Find the equation of the line perpendicular to $y = 2 x - 1$ that passes through the point (4, 3).

1

Identify gradient of given line: m = 2

2

Perpendicular gradient:

m = - \frac{1}{2}

3

Use point-gradient form:

y - 3 = - \frac{1}{2} (x - 4)

4

Expand:

y - 3 = - \frac{1}{2} x + 2

5

Rearrange:

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

Answer: y = -½x + 5

Worked Example04

Find the equation of the line perpendicular to $3 x - 2 y + 8 = 0$ that passes through (6, -1). Give your answer in the form $a x + b y + c = 0$ .

1

Find gradient:

- 2 y = - 3 x - 8

, so

y = \frac{3}{2} x + 4

, m =

\frac{3}{2}

2

Perpendicular gradient:

m = - \frac{2}{3}

3

Use point-gradient form:

y - (- 1) = - \frac{2}{3} (x - 6)

4

y + 1 = - \frac{2}{3} (x - 6)

5

Multiply by 3:

3 y + 3 = - 2 (x - 6) = - 2 x + 12

6

Rearrange:

2 x + 3 y + 3 - 12 = 0

Answer: 2x + 3y - 9 = 0

Classifying Lines

Given two lines, you can determine their relationship by comparing gradients:

Parallel

m_{1} = m_{2}

Equal gradients

Perpendicular

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

Product is -1

Method

Convert both lines to y = mx + c form

Identify both gradients m₁ and m₂

Check: Are they equal? (parallel) Does m₁ × m₂ = -1? (perpendicular)

If neither condition is met → neither parallel nor perpendicular

Worked Example05

Determine whether the lines $y = 4 x + 1$ and $2 y - 8 x + 3 = 0$ are parallel, perpendicular, or neither.

1

Line 1:

y = 4 x + 1

→ gradient m₁ = 4

2

Line 2:

2 y - 8 x + 3 = 0

→

2 y = 8 x - 3

→

y = 4 x - \frac{3}{2}

3

Gradient m₂ = 4

4

Compare: m₁ = m₂ = 4 ✓

Answer: The lines are PARALLEL (equal gradients)

Summary

Key Conditions

RelationshipConditionExample

Parallel

m_{1} = m_{2}

m = 3 and m = 3

Perpendicular

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

m = 2 and m = -½

Exam Strategy

Always find the gradient of the given line first

For parallel: use the same gradient

For perpendicular: use the negative reciprocal

Use point-gradient form with the new point

Convert to the required form at the end

Practice

Question 1: Find the equation of the line parallel to

y = 5 x - 3

that passes through the point (2, 4). Give your answer in the form

y = m x + c

.

1

Given line has gradient m = 5

2

Parallel line has same gradient m = 5

3

Using (2, 4):

y - 4 = 5 (x - 2)

4

y - 4 = 5 x - 10

→

y = 5 x - 6

Answer: y = 5x - 6

Practice

Question 2: Find the equation of the line perpendicular to

y = - 3 x + 7

that passes through the point (6, 1). Give your answer in the form

a x + b y + c = 0

.

1

Given gradient m = -3

2

Perpendicular gradient =

\frac{1}{3}

3

Using (6, 1):

y - 1 = \frac{1}{3} (x - 6)

4

Multiply by 3:

3 y - 3 = x - 6

5

Rearrange:

x - 3 y - 3 = 0

Answer: x - 3y - 3 = 0

Practice

Question 3: Determine whether each pair of lines is parallel, perpendicular, or neither:

(a)

y = 2 x + 5

and

4 x - 2 y + 1 = 0

(b)

y = 3 x - 1

and

x + 3 y - 6 = 0

(a)

•

Line 1: m₁ = 2

•

Line 2:

- 2 y = - 4 x - 1

→

y = 2 x + \frac{1}{2}

→ m₂ = 2

•

m₁ = m₂ → PARALLEL

(b)

•

Line 1: m₁ = 3

•

Line 2:

3 y = - x + 6

→

y = - \frac{1}{3} x + 2

→ m₂ =

- \frac{1}{3}

•

m₁ × m₂ = 3 × (-⅓) = -1 → PERPENDICULAR