Mathematics questions, chapter 10

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Gradient of a Straight Line

43 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

The gradient of a straight line is the ratio of

the horizontal distance to the vertical distance
the vertical distance to the horizontal distance
the length of the line to the horizontal distance
the x-intercept to the y-intercept

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Gradient = vertical distance ÷ horizontal distance (rise over run). Option A is upside down, so a steeper line would give a smaller number.

For the teacher: Remembering, low difficulty, standard 10.1.1

Textbook p. 192

Question 2

Line KL makes an angle of 45° with the x-axis and line MN makes an angle of 30° with the x-axis. Both lines rise to the right. Which statement is true?

KL is steeper than MN, so KL has the greater gradient.
MN is steeper than KL, so MN has the greater gradient.
KL and MN have the same gradient.
MN has the greater gradient because 30 is a smaller number.

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The larger the angle from the x-axis, the steeper the line and the greater the gradient. 45° is larger than 30°, so KL is steeper.

For the teacher: Understanding, low difficulty, standard 10.1.1

Textbook p. 191

Question 3

Diagram 1 shows a ramp AB. What is the gradient of AB?

Right-angled triangle for ramp AB: horizontal distance 12 m, vertical distance 5 m — Diagram 1

125
517
512
1213

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Gradient = vertical ÷ horizontal = 5 ÷ 12. Option A is horizontal ÷ vertical. 17 comes from adding 5 + 12.

For the teacher: Applying, low difficulty, standard 10.1.1

Textbook p. 192

Question 4

What is the gradient of the straight line passing through A (2, 3) and B (6, 11)?

12
4
−2
2

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m = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. Option A is the inverse (4 ÷ 8).

For the teacher: Applying, low difficulty, standard 10.1.2

Textbook p. 194

Question 5

Find the gradient of the straight line passing through P (2, 3) and Q (8, 5).

13
3
45
−13

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m = (5 − 3) ÷ (8 − 2) = 2 ÷ 6 = 1/3. Option B divides change in x by change in y. Option C adds the coordinates instead of subtracting.

For the teacher: Applying, low difficulty, standard 10.1.2

Textbook p. 194

Question 6

A straight line has x-intercept −6 and y-intercept 9. Its gradient is

−32
−23
32
23

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m = −(y-intercept) ÷ (x-intercept) = −9 ÷ (−6) = 3/2. The two minus signs make a positive gradient. Option A forgets the minus sign in the formula.

For the teacher: Applying, low difficulty, standard 10.1.2

Textbook p. 194

Question 7

Which straight line has an undefined gradient?

The line passing through (1, 5) and (6, 5)
The line passing through (2, −3) and (2, 6)
The line passing through (0, 0) and (4, 4)
The line passing through (0, 3) and (−3, 0)

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Both points have x = 2, so the line is parallel to the y-axis and x₂ − x₁ = 0, which cannot be a divisor. Option A is parallel to the x-axis, so its gradient is 0.

For the teacher: Understanding, low difficulty, standard 10.1.3

Textbook p. 196

Question 8

Diagram 2 shows four straight lines P, Q, R and S drawn from the origin. Which line is the steepest?

Four lines from the origin: P to (4, 2), Q to (1, 3), R to (−1, 4), S to (−4, 2) — Diagram 2

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Steepness depends on the absolute value of the gradient. P has m = 1/2, Q has m = 3, R has m = −4 and S has m = −1/2. The greatest absolute value is |−4| = 4, so R is the steepest. Q has the greatest gradient value, but its line is less steep than R.

For the teacher: Analysing, medium difficulty, standard 10.1.3

Textbook p. 196

Question 9

Each step of a staircase is 18 cm high and 24 cm deep. The gradient of the staircase is

43
34
35
1842

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Vertical ÷ horizontal = 18 ÷ 24 = 3/4. Option A is upside down. Option D adds the two distances.

For the teacher: Applying, low difficulty, standard 10.1.4

Textbook p. 197

Question 10

A straight line passes through the point (0, −6) and has a gradient of 3. Its x-intercept is

2
−2
18
−18

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x-intercept = −(y-intercept) ÷ m = −(−6) ÷ 3 = 2. Option B forgets the minus sign in the formula.

For the teacher: Applying, low difficulty, standard 10.1.5

Textbook p. 198

Question 11

A straight line passes through the point (6, 0) and has a gradient of −32. Its y-intercept is

−9
−4
4
9

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y-intercept = −m × (x-intercept) = −(−3/2) × 6 = 9. Option A drops the minus sign in the formula. Options B and C use 6 ÷ (3/2) = 4, which is the wrong rearrangement.

For the teacher: Applying, low difficulty, standard 10.1.5

Textbook p. 199

Question 12

The gradient of the straight line passing through P (2, k) and Q (6, 10) is 32. Find the value of k.

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(10 − k) ÷ (6 − 2) = 3/2, so 10 − k = 6 and k = 4. Option B adds 6 instead of subtracting.

For the teacher: Analysing, medium difficulty, standard 10.1.5

Textbook p. 203

Section B: structured questions

31 marks

Answer all questions. Show your working.

Question 13

Diagram 3 shows two playground slides, P and Q.

Slide P is a right triangle with height 2.4 m and horizontal distance 3.0 m. Slide Q has height 3.0 m and horizontal distance 2.5 m — Diagram 3

[1 mark] (a)Calculate the gradient of slide P.

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2.4 ÷ 3.0 = 0.8 (or 4/5) [1]1

Tick each point you got.

[1 mark] (b)Calculate the gradient of slide Q.

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3.0 ÷ 2.5 = 1.2 (or 6/5) [1]1

Tick each point you got.

[1 mark] (c)Which slide is steeper? Give a reason.

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Slide Q, because its gradient (1.2) is greater than that of P (0.8) [1]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 10.1.1

Textbook p. 192

Question 14

Diagram 4 shows a straight line AB. It cuts the x-axis at A and the y-axis at B.

Line AB through A on the x-axis at 8 and B on the y-axis at −6 — Diagram 4

[2 marks] (a)State the x-intercept and the y-intercept of AB.

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x-intercept = 8 [1]1
y-intercept = −6 [1]1

Tick each point you got.

[2 marks] (b)Calculate the gradient of AB.

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m = −(−6) ÷ 8 [1]1
= 6/8 = 3/4 [1]1

Tick each point you got.

[1 mark] (c)State whether the gradient is positive or negative, and how you can tell from Diagram 4.

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Positive, because the line rises to the right [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 10.1.2

Textbook p. 194

Question 15

Four straight lines are given by two points each: AB through A (1, 2) and B (5, 2), CD through C (3, −1) and D (3, 6), EF through E (0, 0) and F (2, 6), and GH through G (1, 5) and H (4, −1).

[2 marks] (a)Find the gradient of AB and of CD.

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AB: (2 − 2) ÷ (5 − 1) = 0 [1]1
CD: x₂ − x₁ = 0, so the gradient is undefined [1]1

Tick each point you got.

[2 marks] (b)Find the gradient of EF and of GH.

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EF: (6 − 0) ÷ (2 − 0) = 3 [1]1
GH: (−1 − 5) ÷ (4 − 1) = −6 ÷ 3 = −2 [1]1

Tick each point you got.

[1 mark] (c)Arrange the gradients of AB, EF and GH in descending order.

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3, 0, −2 (EF, AB, GH) [1]1

Tick each point you got.

For the teacher: Analysing, low difficulty, standard 10.1.3

Textbook p. 196

Question 16

A straight line passes through P (0, 6) and has a gradient of −32.

[2 marks] (a)Find the x-intercept of the line.

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x-intercept = −6 ÷ (−3/2) [1]1
= 4 [1]1

Tick each point you got.

[2 marks] (b)The line also passes through Q (8, k). Find the value of k.

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(k − 6) ÷ (8 − 0) = −3/2 [1]1
k − 6 = −12, k = −6 [1]1

Check with the intercepts: the line crosses the x-axis at (4, 0), and going 4 more along x drops y by 6, so y = −6.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 10.1.5

Textbook p. 198

Question 17

The vertices of a quadrilateral ABCD are A (−2, 1), B (1, 4), C (5, 2) and D (2, −1).

[2 marks] (a)Calculate the gradients of AB, BC, CD and DA.

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AB = (4 − 1) ÷ (1 + 2) = 1 and CD = (−1 − 2) ÷ (2 − 5) = 1 [1]1
BC = (2 − 4) ÷ (5 − 1) = −1/2 and DA = (1 + 1) ÷ (−2 − 2) = −1/2 [1]1

Tick each point you got.

[2 marks] (b)Name the type of quadrilateral and give a reason.

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Parallelogram [1]1
Opposite sides have equal gradients, so they are parallel (AB // DC and BC // AD) [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 10.1.5

Textbook p. 199

Question 18

Amir wants the gradient of the line through P (2, 5) and Q (6, −3). His working is: m = (6 − 2) ÷ (−3 − 5) = 4 ÷ (−8) = −1/2.

[1 mark] (a)Is Amir's answer correct? State his mistake.

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No. He divided the change in x by the change in y (he used x over y) [1]1

Tick each point you got.

[2 marks] (b)Show the correct working and answer.

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m = (−3 − 5) ÷ (6 − 2) [1]1
= −8 ÷ 4 = −2 [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 10.1.2

Textbook p. 194

Question 19

A is the point (2, 3). Write the coordinates of a point B, different from A, so that the line AB has

[1 mark] (a)a gradient of 2.

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(3, 5): m = (5 − 3) ÷ (3 − 2) = 2

Any point with (y − 3) ÷ (x − 2) = 2, e.g. (3, 5), (4, 7), (0, −1) [1]1

Tick each point you got.

[1 mark] (b)a gradient of zero.

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(7, 3): the line is parallel to the x-axis

Any point with y = 3 (x ≠ 2), e.g. (7, 3) [1]1

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[1 mark] (c)an undefined gradient.

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(2, 8): the line is parallel to the y-axis

Any point with x = 2 (y ≠ 3), e.g. (2, 8) [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 10.1.3

Textbook p. 196

Question 20

Diagram 5 shows the roof of a house. The roof is an isosceles triangle with a width of 24 m and a height of 5 m.

Isosceles triangle roof with a base of 24 m and a vertical height of 5 m drawn to the middle of the base — Diagram 5

[2 marks] (a)Calculate the gradient of the roof.

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Horizontal distance = 24 ÷ 2 = 12 m [1]1
m = 5 ÷ 12 = 5/12 [1]1

Tick each point you got.

[2 marks] (b)Calculate the length of one slanting side of the roof.

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√(5² + 12²) = √169 [1]1
= 13 m [1]1

Pythagoras' theorem: slant² = vertical² + horizontal². 5, 12 and 13 form a Pythagorean triple.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 10.1.5

Textbook p. 203

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