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GCSE · Algebra: straight-line graphs

5.14D Gradient of a Line: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculators not allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Find the gradient of the straight line joining the points (2, 3) and (5, 12).

(Total for Question 1 is 1 mark)

Find the gradient of the straight line joining the points (0, 7) and (4, -1).

(Total for Question 2 is 1 mark)

Find the gradient of the straight line joining the points (-3, -4) and (1, -4).

(Total for Question 3 is 1 mark)

Find the gradient of the straight line joining the points (5, -2) and (5, 6).

(Total for Question 4 is 1 mark)

Write down the gradient of the line with equation y = 7x + 2.

(Total for Question 5 is 1 mark)

Write down the gradient of the line with equation y = -4x - 9.

(Total for Question 6 is 1 mark)

Write down the gradient of the line with equation y = 6 - x.

(Total for Question 7 is 1 mark)

Rearrange 2y = 8x - 6 into the form y = mx + c and state the gradient.

(Total for Question 8 is 2 marks)

Rearrange 3x + y = 11 into the form y = mx + c and state the gradient and the y-intercept.

(Total for Question 9 is 2 marks)

Rearrange 5x - 2y = 12 into the form y = mx + c and state the gradient.

(Total for Question 10 is 2 marks)

Find the gradient of the straight line joining the points (1, 1) and (4, 7).

(Total for Question 11 is 2 marks)

Find the gradient of the straight line joining the points (2, 5) and (8, 1).

(Total for Question 12 is 2 marks)

A line L1 has gradient 3. Line L2 is parallel to L1 and passes through a different point. State the gradient of L2, giving a reason for your answer.

(Total for Question 13 is 2 marks)

A line has gradient 6. Find the gradient of a line perpendicular to it.

(Total for Question 14 is 2 marks)

A line has gradient -3/5. Find the gradient of a line perpendicular to it.

(Total for Question 15 is 2 marks)

Find the gradient of the straight line joining the points (-1, 6) and (3, -2).

(Total for Question 16 is 2 marks)

A straight line passes through the points A(-1, 4) and B(3, 12).

(a)Find the gradient of the line AB.(2)

(b)Hence find the equation of the line AB, giving your answer in the form y = mx + c.(2)

(Total for Question 17 is 4 marks)

The points P(2, 3), Q(5, 15) and R(9, k) lie on the same straight line. Find the value of k, showing your working.

(Total for Question 18 is 4 marks)

A narrowboat travels along a canal. After t hours, its distance from the lock, d km, is given by d = 2.5t + 3 for 0 ≤ t ≤ 4.

(a)State the gradient of this line and explain what it represents in this context.(2)

(b)State the value of the d-intercept and explain what it means in this context.(2)

(Total for Question 19 is 4 marks)

Line L1 has equation y = 3x - 2. Line L2 is perpendicular to L1 and passes through the point (0, -4). Find the equation of L2, giving your answer in the form y = mx + c.

(Total for Question 20 is 3 marks)

Line L1 passes through the points (1, 3) and (3, 9). Line L2 passes through the points (2, 10) and (5, 1). Determine, showing your working, whether L1 and L2 are perpendicular.

(Total for Question 21 is 4 marks)

A quadrilateral has vertices at A(0, 0), B(4, 2), C(6, -2) and D(2, -4). Show that AB is parallel to DC.

(Total for Question 22 is 3 marks)

A line L passes through the point (3, -2) and has the same gradient as the line 5x + y = 4. Find the equation of L, giving your answer in the form y = mx + c.

(Total for Question 23 is 3 marks)

Line L1 passes through the points A(2, 5) and B(8, k). Line L2 has equation 3x + y = 12 and is perpendicular to L1. Find the value of k.

(Total for Question 24 is 4 marks)

A straight line has equation ay + bx = c, where a, b and c are non-zero constants.

(a)Show that the gradient of this line is -b/a.(2)

(b)A second line has equation 3y - 9x = 5. Given that the line ay + bx = c is perpendicular to this second line, and that a = 4, find the value of b.(3)

(Total for Question 25 is 5 marks)

Mark scheme · 5.14D Gradient of a Line: Fluency and Exam Drill

Question 1

B1 3 cao
Answer: 3

Question 2

B1 -2 cao
Answer: -2

Question 3

B1 0 cao
Answer: 0

Question 4

B1 undefined (the line is vertical), oe
Answer: Undefined (vertical line)

Question 5

B1 7 cao
Answer: 7

Question 6

B1 -4 cao
Answer: -4

Question 7

B1 -1 cao
Answer: -1

Question 8

M1 divides every term by 2, oe
A1 4 cao (accept full statement y = 4x - 3)
Answer: 4

Question 9

M1 subtracts 3x from both sides, oe
A1 gradient -3 and y-intercept 11 both stated correctly, cao
Answer: Gradient = -3, y-intercept = 11

Question 10

M1 rearranges to isolate y correctly, oe
A1 5/2 oe (accept 2.5) cao
Answer: 5/2

Question 11

M1 (7 - 1)/(4 - 1), oe substitution into the gradient formula
A1 2 cao
Answer: 2

Question 12

M1 (1 - 5)/(8 - 2), oe substitution into the gradient formula
A1 -2/3 oe cao
Answer: -2/3

Question 13

B1 3 cao
B1 correct reason given, e.g. parallel lines have the same (equal) gradient
Answer: 3, because parallel lines have equal gradients

Question 14

M1 uses (gradient) x (perpendicular gradient) = -1, oe
A1 -1/6 oe cao
Answer: -1/6

Question 15

M1 uses (gradient) x (perpendicular gradient) = -1, oe, taking the negative reciprocal of -3/5
A1 5/3 oe cao
Answer: 5/3

Question 16

M1 (-2 - 6)/(3 - (-1)), oe substitution into the gradient formula with correct signs
A1 -2 cao
Answer: -2

Question 17

(a) M1 (12 - 4)/(3 - (-1)), oe
(a) A1 2 cao
(a) Answer: 2
(b) M1 uses y - 4 = 2(x + 1), oe, or substitutes A or B into y = 2x + c to find c
(b) A1 y = 2x + 6 cao
(b) Answer: y = 2x + 6

Question 18

B1 states or uses that points on the same straight line have equal gradients between any pair of them
M1 gradient of PQ = (15 - 3)/(5 - 2) = 4
M1 sets up (k - 15)/(9 - 5) = 4, oe (or the equivalent equation using PR), ft their gradient
A1 k = 31 cao
Answer: k = 31

Question 19

(a) B1 2.5 cao
(a) B1 correct interpretation, e.g. the speed of the narrowboat is 2.5 km/h
(a) Answer: 2.5, the speed of the narrowboat in km/h
(b) B1 3 cao
(b) B1 correct interpretation, e.g. the narrowboat is already 3 km from the lock when t = 0 (the start of the recorded journey)
(b) Answer: 3, the narrowboat is 3 km from the lock at t = 0

Question 20

M1 identifies the gradient of L1 as 3
M1 uses the negative reciprocal to find the gradient of L2 as -1/3, ft
A1 y = -1/3 x - 4 cao
Answer: y = -1/3 x - 4

Question 21

M1 gradient of L1 = (9 - 3)/(3 - 1) = 3
M1 gradient of L2 = (1 - 10)/(5 - 2) = -3
M1 tests the product of the two gradients, ft their gradients
A1 correct conclusion: not perpendicular, since 3 x (-3) = -9, not -1, cao
Answer: L1 and L2 are not perpendicular

Question 22

M1 gradient of AB = (2 - 0)/(4 - 0) = 1/2
M1 gradient of DC = (-2 - (-4))/(6 - 2) = 1/2
A1 correct conclusion with reason: gradients are equal (both 1/2), so AB is parallel to DC, cso
Answer: Gradient of AB = gradient of DC = 1/2, so AB is parallel to DC

Question 23

M1 rearranges 5x + y = 4 to find gradient -5
M1 uses y - (-2) = -5(x - 3), oe, or substitutes (3, -2) into y = -5x + c to find c
A1 y = -5x + 13 cao
Answer: y = -5x + 13

Question 24

M1 rearranges 3x + y = 12 to find the gradient of L2 as -3
M1 uses the perpendicular condition to find the gradient of L1 as 1/3, ft their gradient of L2
M1 sets up (k - 5)/(8 - 2) = 1/3, oe, ft their gradient of L1
A1 k = 7 cao
Answer: k = 7

Question 25

(a) M1 rearranges ay + bx = c to make y the subject: ay = -bx + c, then y = (-b/a)x + c/a
(a) A1 correctly identifies the coefficient of x as the gradient -b/a, cso
(a) Answer: Gradient = -b/a
(b) M1 rearranges 3y - 9x = 5 to find its gradient = 3
(b) M1 uses the perpendicular condition (-b/a)(3) = -1, ft their expression -b/a from (a) and their gradient of the second line
(b) A1 b = 4/3 oe cao
(b) Answer: b = 4/3

Gradient of a Line: Fluency and Exam Drill - Worksheets, Questions and Revision