The diagram shows a straight line drawn on a grid.
(Total for Question 1 is 2 marks)
2
The diagrams show two straight lines drawn on grids.
(a)Write down the gradient of the horizontal line shown in diagram (a).(1)
(b)Write down the gradient of the vertical line shown in diagram (b).(1)
(Total for Question 2 is 2 marks)
3
For each equation, write down the gradient and the y-intercept of the line.
(a)y = 5x - 2(1)
(b)y = -3x + 7(1)
(Total for Question 3 is 2 marks)
4
The table shows coordinates of points that lie on a straight line. x: 0, 1, 2, 3 y: 5, 8, 11, 14 Find the gradient of the line.
(Total for Question 4 is 2 marks)
5
Calculate the gradient of the line joining the points A(1, 2) and B(5, 10).
(Total for Question 5 is 2 marks)
6
Calculate the gradient of the line joining the points P(-3, 5) and Q(2, -5).
(Total for Question 6 is 2 marks)
7
A straight line has equation 2x + y = 8.
(a)Rearrange the equation into the form y = mx + c.(2)
(b)Write down the gradient and the y-intercept of the line.(2)
(Total for Question 7 is 4 marks)
8
The graph shows the cost, C pounds, of hiring a bicycle for t hours.
(a)Calculate the gradient of the line.(2)
(b)State what the gradient represents in this context.(1)
(Total for Question 8 is 3 marks)
9
The graph shows the distance, d kilometres, travelled by a train, t hours after leaving a station.
(a)Calculate the gradient of the line.(2)
(b)State what the gradient represents, including units.(1)
(Total for Question 9 is 3 marks)
10
Line L1 has equation y = 4x - 3. Line L2 is parallel to L1 and passes through the point (0, 5). Find an equation for L2.
(Total for Question 10 is 3 marks)
11
Find an equation of the straight line that passes through the points (2, 1) and (6, 9). Give your answer in the form y = mx + c.
(Total for Question 11 is 4 marks)
12
A straight line has gradient 5 and passes through the points (3, k) and (5, 16). Find the value of k.
(Total for Question 12 is 3 marks)
13
Show that the line with equation 3y = 6x + 9 is parallel to the line with equation y = 2x - 1.
(Total for Question 13 is 2 marks)
14
A straight line has gradient -3 and passes through the point (4, 2). Find an equation of the line. Give your answer in the form y = mx + c.
(Total for Question 14 is 3 marks)
15
Which of these lines is the steepest?
A) y = 2x + 1
B) y = -5x + 3
C) y = 0.5x - 2
D) y = 3x
(Total for Question 15 is 1 mark)
16
Line L has equation y = (1/2)x + 3.
(a)Write down the gradient of a line that is perpendicular to L.(1)
(b)Find an equation of the line that is perpendicular to L and passes through the point (4, 1). Give your answer in the form y = mx + c.(3)
(Total for Question 16 is 4 marks)
17
Use gradients to answer each part below.
(a)The points A(1, 2), B(4, 11) and C(k, 29) lie on a straight line. Find the value of k.(3)
(b)The line y = mx - 7 is perpendicular to the line y = -(1/3)x + 2. Find the value of m.(2)
(Total for Question 17 is 5 marks)
18
The points A(-2, 5) and B(6, 1) are joined to form a line segment. Find an equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.
(Total for Question 18 is 5 marks)
19
A straight line passes through the point (2, -1) and is parallel to the line with equation 4x - 2y = 6. Find an equation of the line. Give your answer in the form ax + by = c, where a, b and c are integers.
(Total for Question 19 is 4 marks)
Mark scheme · 5.14 Gradient of a Line
Question 1
M1 correct method to find rise and run using two points on the line, e.g. sight of 8 and 4, or 8/4
A1 2 cao
Answer: Gradient = 2
Question 2
(a) B1 0 cao
(a) Answer: 0
(b) B1 undefined oe (e.g. no gradient, infinite)
(b) Answer: Undefined (the line has no defined gradient)
Question 3
(a) B1 gradient = 5 and y-intercept = (0, -2), both correct
(a) Answer: Gradient = 5, y-intercept = (0, -2)
(b) B1 gradient = -3 and y-intercept = (0, 7), both correct
(b) Answer: Gradient = -3, y-intercept = (0, 7)
Question 4
M1 identifies constant increase of 3 in y for an increase of 1 in x, oe (e.g. uses two points in gradient formula)
A1 3 cao
Answer: Gradient = 3
Question 5
M1 (10 - 2) / (5 - 1) oe
A1 2 cao
Answer: Gradient = 2
Question 6
M1 (-5 - 5) / (2 - (-3)) oe
A1 -2 cao
Answer: Gradient = -2
Question 7
(a) M1 subtracts 2x from both sides, e.g. y = 8 - 2x oe
(a) A1 y = -2x + 8 cao
(a) Answer: y = -2x + 8
(b) B1 gradient = -2 (ft from part a)
(b) B1 y-intercept = (0, 8) (ft from part a)
(b) Answer: Gradient = -2, y-intercept = (0, 8)
Question 8
(a) M1 (17 - 5) / (4 - 0) oe
(a) A1 3 cao
(a) Answer: Gradient = 3
(b) B1 oe, e.g. the hire cost increases by 3 pounds for each extra hour / the hourly hire rate is 3 pounds per hour
(b) Answer: The cost of hiring the bicycle increases by 3 pounds for each extra hour (the hire rate is 3 pounds per hour).
Question 9
(a) M1 (170 - 0) / (2 - 0) oe
(a) A1 85 cao
(a) Answer: Gradient = 85
(b) B1 oe, e.g. the speed of the train is 85 km/h
(b) Answer: The gradient represents the speed of the train, which is 85 km/h.
Question 10
B1 gradient of L2 = 4, since parallel lines have equal gradients
M1 uses gradient 4 with the point (0, 5), e.g. y = 4x + c, 5 = 4(0) + c
A1 y = 4x + 5 cao
Answer: y = 4x + 5
Question 11
M1 (9 - 1) / (6 - 2) oe
A1 gradient = 2
M1 substitutes a point into y = 2x + c, e.g. 1 = 2(2) + c, or uses y - 1 = 2(x - 2) (dep)
A1 y = 2x - 3 cao
Answer: y = 2x - 3
Question 12
M1 (16 - k) / (5 - 3) = 5 oe
M1 16 - k = 10 (dep)
A1 k = 6 cao
Answer: k = 6
Question 13
M1 rearranges 3y = 6x + 9 to y = 2x + 3 oe
A1 cso: states both lines have gradient 2, so they are parallel
Answer: Both lines have gradient 2, so they are parallel.
Question 14
M1 y - 2 = -3(x - 4) oe, or 2 = -3(4) + c
M1 correctly expands/rearranges (dep)
A1 y = -3x + 14 cao
Answer: y = -3x + 14
Question 15
B1 B cao
Answer: B) y = -5x + 3
Question 16
(a) B1 -2 cao
(a) Answer: -2
(b) M1 uses gradient -2 (ft from part a) with the point (4, 1), e.g. 1 = -2(4) + c
(b) M1 solves for c correctly (dep)
(b) A1 y = -2x + 9 cao
(b) Answer: y = -2x + 9
Question 17
(a) M1 finds gradient of AB = (11 - 2) / (4 - 1) = 3
(a) M1 sets up equation using gradient AC = 3, e.g. (29 - 2) / (k - 1) = 3 oe (dep)
(a) A1 k = 10 cao
(a) Answer: k = 10
(b) M1 m x (-1/3) = -1 oe
(b) A1 m = 3 cao
(b) Answer: m = 3
Question 18
M1 finds midpoint of AB, e.g. ((-2 + 6)/2, (5 + 1)/2) oe
A1 midpoint = (2, 3)
M1 finds gradient of AB = (1 - 5) / (6 - (-2)) = -1/2 oe
M1 uses perpendicular gradient = 2 (negative reciprocal, ft their gradient) with the midpoint (dep)
A1 y = 2x - 1 cao
Answer: y = 2x - 1
Question 19
M1 rearranges 4x - 2y = 6 to find the gradient, e.g. y = 2x - 3, gradient = 2
M1 uses gradient 2 with the point (2, -1), e.g. y - (-1) = 2(x - 2) oe (dep)
M1 simplifies to y = 2x - 5, ft their gradient (dep)