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Amy Ferguson Moncure

IB AA · 4A Equations of Straight Lines

Parallel and Perpendicular Lines

Match a gradient, or use the negative reciprocal.

Jump to section:

parallel: same m

perpendicular: m₁ m₂ = −1

What to Remember

Parallel

Parallel lines have the same gradient.

Perpendicular

Perpendicular gradients multiply to −1. If one gradient is m, the other is −1/m.

Horizontal and vertical

Gradient 0 is horizontal. The perpendicular line is vertical, x = a constant.

Then the point

Use y − y₁ = m(x − x₁) with the new gradient and the given point.

Easy

Easy

Parallel to a line in the form y = mx + c.

Example
Find the line parallel to y = −2x + 5 through (3, 1).

y = −2x + 5

A parallel line has the same gradient, m = −2.

y − y₁ = m(x − x₁)

y − 1 = −2(x − 3)

y − 1 = −2x + 6

y = −2x + 7

Answer: y = −2x + 7

1
Find the equation of the line parallel to y = 3x + 2 through (1, 4).
Show Answer

y = 3x + 2

A parallel line has the same gradient, m = 3.

y − y₁ = m(x − x₁)

y − 4 = 3(x − 1)

y − 4 = 3x − 3

y = 3x + 1

Answer: y = 3x + 1

2
Find the equation of the line parallel to y = −x + 1 through (0, 5).
Show Answer

y = −x + 1

A parallel line has the same gradient, m = −1.

y − y₁ = m(x − x₁)

y − 5 = −1(x − 0)

y − 5 = −x

y = −x + 5

Answer: y = −x + 5

3
Find the equation of the line parallel to y = 2x − 6 through (−1, 3).
Show Answer

y = 2x − 6

A parallel line has the same gradient, m = 2.

y − y₁ = m(x − x₁)

y − 3 = 2(x − (−1))

y − 3 = 2x + 2

y = 2x + 5

Answer: y = 2x + 5

4
Find the equation of the line parallel to y = 12x + 1 through (2, 5).
Show Answer

y = 12x + 1

A parallel line has the same gradient, m = 12.

y − y₁ = m(x − x₁)

y − 5 = 12(x − 2)

y − 5 = 12x − 1

y = 12x + 4

Answer: y = 12x + 4

5
Find the equation of the line parallel to y = 3 through (4, −2).
Show Answer

y = 3

A parallel line has the same gradient, m = 0.

y − y₁ = m(x − x₁)

y − (−2) = 0(x − 4)

y − (−2) = 0

y = −2

Answer: y = −2

Medium

Medium

Perpendicular to y = mx + c.

Example
Find the line perpendicular to y = 4x − 1 through (2, 5).

y = 4x − 1

Perpendicular gradients multiply to −1.

m₂ = −14

m₂ = −14

y − y₁ = m(x − x₁)

y − 5 = −14(x − 2)

y − 5 = −14x + 12

y = −14x + 112

Answer: y = −14x + 112

1
Find the equation of the line perpendicular to y = 2x − 1 through (4, 3).
Show Answer

y = 2x − 1

Perpendicular gradients multiply to −1.

m₂ = −12

m₂ = −12

y − y₁ = m(x − x₁)

y − 3 = −12(x − 4)

y − 3 = −12x + 2

y = −12x + 5

Answer: y = −12x + 5

2
Find the equation of the line perpendicular to y = −3x + 4 through (1, 5).
Show Answer

y = −3x + 4

Perpendicular gradients multiply to −1.

m₂ = −1−3

m₂ = 13

y − y₁ = m(x − x₁)

y − 5 = 13(x − 1)

y − 5 = 13x − 13

y = 13x + 143

Answer: y = 13x + 143

3
Find the equation of the line perpendicular to y = 12x − 4 through (3, 2).
Show Answer

y = 12x − 4

Perpendicular gradients multiply to −1.

m₂ = −112

m₂ = −2

y − y₁ = m(x − x₁)

y − 2 = −2(x − 3)

y − 2 = −2x + 6

y = −2x + 8

Answer: y = −2x + 8

4
Find the equation of the line perpendicular to y = −23x + 5 through (0, 1).
Show Answer

y = −23x + 5

Perpendicular gradients multiply to −1.

m₂ = −1−23

m₂ = 32

y − y₁ = m(x − x₁)

y − 1 = 32(x − 0)

y − 1 = 32x

y = 32x + 1

Answer: y = 32x + 1

5
Find the equation of the line perpendicular to y = x − 3 through (−2, 4).
Show Answer

y = x − 3

Perpendicular gradients multiply to −1.

m₂ = −11

m₂ = −1

y − y₁ = m(x − x₁)

y − 4 = −1(x − (−2))

y − 4 = −x − 2

y = −x + 2

Answer: y = −x + 2

Hard

Hard

Perpendicular to ax + by + c = 0.

Example
Find the line perpendicular to 2x − y − 6 = 0 through (4, 1).

2x − y − 6 = 0

y = 2x − 6

The gradient of the given line is 2.

Perpendicular gradients multiply to −1.

m₂ = −12

m₂ = −12

y − y₁ = m(x − x₁)

y − 1 = −12(x − 4)

y − 1 = −12x + 2

y = −12x + 3

Answer: y = −12x + 3

1
Find the line perpendicular to 2x − 4y + 5 = 0 through (1, 2).
Show Answer

2x − 4y + 5 = 0

y = 12x + 54

The gradient of the given line is 12.

Perpendicular gradients multiply to −1.

m₂ = −112

m₂ = −2

y − y₁ = m(x − x₁)

y − 2 = −2(x − 1)

y − 2 = −2x + 2

y = −2x + 4

Answer: y = −2x + 4

2
Find the line perpendicular to 3x + y − 6 = 0 through (2, 4).
Show Answer

3x + y − 6 = 0

y = −3x + 6

The gradient of the given line is −3.

Perpendicular gradients multiply to −1.

m₂ = −1−3

m₂ = 13

y − y₁ = m(x − x₁)

y − 4 = 13(x − 2)

y − 4 = 13x − 23

y = 13x + 103

Answer: y = 13x + 103

3
Find the line perpendicular to x − 2y + 8 = 0 through (0, 3).
Show Answer

x − 2y + 8 = 0

y = 12x + 4

The gradient of the given line is 12.

Perpendicular gradients multiply to −1.

m₂ = −112

m₂ = −2

y − y₁ = m(x − x₁)

y − 3 = −2(x − 0)

y − 3 = −2x

y = −2x + 3

Answer: y = −2x + 3

4
Find the line perpendicular to 2x + y − 6 = 0 through (−1, 5).
Show Answer

2x + y − 6 = 0

y = −2x + 6

The gradient of the given line is −2.

Perpendicular gradients multiply to −1.

m₂ = −1−2

m₂ = 12

y − y₁ = m(x − x₁)

y − 5 = 12(x − (−1))

y − 5 = 12x + 12

y = 12x + 112

Answer: y = 12x + 112

5
Find the line perpendicular to y − 3 = 0 through (2, 6).
Show Answer

y − 3 = 0

y = 3

The gradient of the given line is 0.

A horizontal line has gradient 0.

A perpendicular line is vertical.

x = 2

Answer: x = 2

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