Parallel and perpendicular Quiz (Grade 6)

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Parallel and perpendicular — Notes & Worked Examples Show / Hide

Explanation:

Always rearrange any straight-line equation into the form y = mx + c first so you can read its gradient m. Parallel lines have the exact same gradient (m₁ = m₂).

Perpendicular lines meet at 90° and their gradients are negative reciprocals of each other: m₂ = −1m₁ (flip the fraction upside down and change the sign, so m₁ × m₂ = −1).

Example 1: Find the equation of the line parallel to 2y − 6x = 5 that passes through (1, −2).

Solution:

Rearrange: 2y = 6x + 5 → y = 3x + 2.5, so m = 3

Substitute m = 3 and (1, −2) into y = mx + c:

−2 = 3(1) + c → c = −5

Answer: y = 3x − 5

Example 2: Find the equation of the line perpendicular to y = 23x + 4 passing through (4, −1).

Solution:

Perpendicular gradient m_⊥ = −32 (negative reciprocal)

Sub m = −32 and (4, −1): −1 = −32(4) + c → c = 5

Answer: y = −32x + 5

One Page Guide (PDF) Parallel and perpendicular Worksheets

Question 1

[1 mark]

Find the equation of the line perpendicular to y = 3x - 4 that passes through the point (6, 6).

y =

Worked Solution & Answer

Perpendicular gradient m_⊥ = -13
Substituting (6, 6) gives c = 8. B1

Answer: y = y = -13x + 8

Question 2

[1 mark]

Write down the equation of a line parallel to:
y = 2x + 4

Worked Solution & Answer

Equation of a line: y = mx + c
Parallel lines have the same gradient (m).
e.g. y = 2x + 11

Answer: e.g. y = 2x + 11

Question 3 AQA Nov 2019 1H Q22

[3 marks]

(-3, 10) is a point on line L.
(4, 0) and (6, 10) are points on line M.
L and M are parallel.

Work out the equation of line L.
Give your answer in the form y = mx + c

Answer

Worked Solution & Answer

Gradient of M = 10 - 06 - 4 = 102 = 5 M1
Since L is parallel to M, gradient of L is also 5.
Equation of L using (-3, 10): M1
y - 10 = 5(x - (-3)) = 5(x + 3) = 5x + 15
y = 5x + 25. A1

Answer: Answer y = 5x + 25

Question Paper (PDF) • Mark Scheme (PDF)

Question 4 Practice Set 6 2H Q20

[3 marks]

The line N is drawn below.

Find an equation of the line perpendicular to line N that passes through the point (0, 1).

y =

Worked Solution & Answer

Line N passes through (0, –2) and (2, 4), gradient = 3. Perpendicular gradient = –1/3. Passes through (0, 1) M2
⇒ y = –1/3 x + 1. A1

Answer: y = y = -13x + 1

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[4 marks]

ABCD is a rhombus.
The coordinates of A are (5, 11)
The equation of the diagonal DB is y = 12x + 6
Find an equation of the diagonal AC.

y =

Worked Solution & Answer

Diagonals of a rhombus are perpendicular M1
Gradient of DB = 1/2, so gradient of AC = –1 ÷ (1/2) = –2. Equation of AC is y = –2x + c M1
Substitute A(5, 11): 11 = –2(5) + c ⇒ c = 21 M1
So y = –2x + 21. A1

Answer: y = y = -2x + 21

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 Practice Set 1 1H Q20

[4 marks]

In the diagram A is the point (–2, 0)
B is the point (0, 4)
C is the point (5, –1)
Find an equation of the line that passes through C and is perpendicular to AB.

Worked Solution & Answer

Gradient of AB = (4 − 0)/(0 − (−2)) = 2. Perpendicular gradient = −1/2. Through (5, −1): y − (−1) = −½(x − 5) M3
⇒ y = −½x + 1.5 A1

Answer: y = -12x + 32 (or x + 2y = 3)

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 Specimen Set 1 1H Q23

[4 marks]

A(–2, 0), B(0, 4), C(5, –1). Find an equation of the line that passes through C and is perpendicular to AB.

Find an equation of the line that passes through C and is perpendicular to AB.

y =

Worked Solution & Answer

Gradient of AB = (4 – 0) / (0 – (–2)) = 4/2 = 2. Perpendicular gradient = –1/2. Line through C(5, –1): y – (–1) = –½(x – 5) M3
⇒ y = –½x + 3/2 (or x + 2y = 3). A1

Answer: y = y = -12x + 32

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[5 marks]

In the diagram
A is the point (0, 8)
B is the point (16, 0)
The point D divides the line segment AB in the ratio 1 : 3
The line **L** passes through D.
The gradient of **L** is √(3)
**L** passes through the point with coordinates (-2, f)

Show that f < -4

Worked Solution & Answer

D divides AB in ratio 1 : 3 ⇒ D = (0 + 1/4 × 16, 8 + 1/4 × (0 – 8)) = (4, 6) M2
Equation of L: y – 6 = √3(x – 4) M1
Substitute (–2, f): f – 6 = √3(–2 – 4) = –6√3 ⇒ f = 6 – 6√3 ≈ 6 – 10.392 = –4.392 M1
Since –4.39 < –4, shown. C1

Answer: Shown (f = 6 – 6√3 ≈ –4.39 < –4)

Question Paper (PDF) • Mark Scheme (PDF)

Question 9

[5 marks]

OBCD is a rectangle.
DCE is a straight line.

B has coordinates (2, -4)
E has coordinates (12, -6.5)
Work out the coordinates of D.
You must show all your working.

Worked Solution & Answer

Gradient of OB = –4/2 = –2. Since DCE is parallel to OB, gradient of DE = –2 ⇒ equation of DE is y – (–6.5) = –2(x – 12) ⇒ y = –2x + 17.5. Since OD ⊥ OB, gradient of OD = 1/2 ⇒ equation of OD is y = 0.5x M4
At D: 0.5x = –2x + 17.5 ⇒ 2.5x = 17.5 ⇒ x = 7, y = 3.5 A1

Answer: 7 , 3.5

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Jan 2020 2H Q19

[5 marks]

(a) Equation of line parallel to y = 7 − 4x (1) (b) Line perpendicular to L through (−3, 1) and (2, −2) passing through (−6, 4) in form ax + by + c = 0 (4)

(a) Write down an equation of a line that is parallel to the line with equation y = 7 - 4x [1 mark]

y =

(b) Find an equation of the line that is perpendicular to L and passes through the point with coordinates (-6, 4)
Give your answer in the form ax + by + c = 0 where a, b and c are integers. [4 marks]

Worked Solution & Answer

(a)

Any line with gradient −4, e.g. y = −4x B1

Answer (a): y = y = -4x

(b)

m_L = (−2 − 1)/(2 − (−3)) = −3/5 ⇒ m_perp = 5/3 M3
y − 4 = (5/3)(x + 6) ⇒ 3y − 12 = 5x + 30 ⇒ 5x − 3y + 42 = 0 A1

Answer (b): 5x - 3y + 42 = 0

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 35 marks)

Parallel and perpendicular Worksheets Next: Inequalities on graphs →

Grade 6 Quizzes

14 Topics

6A - Recurring decimals	6B - Harder indices
6C - The product rule	6D - Repeated change
6E - Triple brackets	6F - Parallel and perpendicular
6G - Inequalities on graphs	6H - Similar shapes (area and vol)
6J - Negative scale factors	6K - Circle theorems
6L - Cumulative frequency	6M - Box plots
6N - Combined transformations	6P - Capture recapture