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3.15H Drawing Linear Graphs: Higher Tier Practice

EDEXCEL 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Plot the point with coordinates (3, -2) on a pair of axes.

(Total for Question 1 is 1 mark)

Write down the equation of the straight line passing through the points (0, 5) and (2, 1).

(Total for Question 2 is 2 marks)

A line has equation y = 3x - 4.
Work out the coordinates of the point where this line crosses the y-axis and the x-axis.

(a)Work out the y-intercept.(1)

(b)Work out the x-intercept.(2)

(Total for Question 3 is 3 marks)

The graph of a line through (1, 2) has gradient -1/2.
Find the equation of this line in the form y = mx + c.

(Total for Question 4 is 3 marks)

Two lines L1 and L2 are given by L1: y = 2x + 1 and L2: y = -x + 7.
Find the coordinates of their point of intersection.

(Total for Question 5 is 4 marks)

The line L has equation y = ax + 4 and passes through the point (6, -2).
Find the value of a. Then state the equation of the line perpendicular to L that passes through (6, -2).

(a)Find the value of a.(2)

(b)State the equation of the line perpendicular to L through (6, -2).(3)

(Total for Question 6 is 5 marks)

A company charges a fixed booking fee and a constant hourly rate for hiring equipment. A graph of cost C (pounds) against hours h is a straight line. The graph passes through (0, 18) and (5, 58).
(a) Find the booking fee and the hourly rate. (b) Write down the equation connecting C and h.

(a)Find the booking fee and the hourly rate.(3)

(b)Write down the equation connecting C and h.(3)

(Total for Question 7 is 6 marks)

Show that the family of lines with equation y = mx + (3 - 2m) all pass through a single fixed point independent of m. Find the coordinates of that point.

(a)Show that all lines of the form y = mx + (3 - 2m) pass through a fixed point, and find its coordinates.(8)

(Total for Question 8 is 8 marks)

A cyclist travels at a steady speed along a straight path. Her distance d (metres) from the start after t minutes is modelled by d = 350t + 120.
(a) Interpret the numbers 350 and 120 in this context. (b) Another cyclist starts at the same start but 6 minutes later. For t ≥ 6, her distance is modelled by d = 450(t - 6), where t is minutes since the original start. Find the time (in minutes since the original start) when the second cyclist catches the first.

(a)Interpret the numbers 350 and 120 in context.(2)

(b)Find the time since the original start when the second cyclist catches the first.(8)

(Total for Question 9 is 10 marks)

Given the line L1: 4x - 2y + 6 = 0, (a) rearrange to the form y = mx + c. (b) Sketch L1 and the line L3 which is the image of L1 after a translation of 3 units right and 2 units up. Give the equation of L3.

(a)Rearrange 4x - 2y + 6 = 0 to y = mx + c.(3)

(b)Give the equation of L3, the image of L1 under translation 3 units right and 2 units up. Include a brief sketch showing both lines labelled.(5)

(Total for Question 10 is 8 marks)

Mark scheme · 3.15H Drawing Linear Graphs: Higher Tier Practice

Question 1

B1 point plotted at (3, -2) within tolerance (correct quadrant and position) oe
Answer: (3, -2)

Question 2

M1 find gradient m = (1 - 5) / (2 - 0) = -4/2 = -2 or equivalent method
A1 y = -2x + 5 cao
Answer: y = -2x + 5

Question 3

(a) B1 y-intercept (0, -4) cao
(a) Answer: (0, -4)
(b) M1 set y = 0 and solve 0 = 3x - 4, showing method
(b) A1 x = 4/3, so x-intercept (4/3, 0) cao
(b) Answer: (4/3, 0)

Question 4

M1 substitute point into y = (-1/2)x + c and solve for c: 2 = (-1/2)(1) + c
A1 c = 5/2 or 2.5
A1 y = -1/2 x + 5/2 cao
Answer: y = -1/2 x + 5/2

Question 5

M1 set 2x + 1 = -x + 7 and form equation 3x = 6 or equivalent
A1 x = 2 cao
M1 substitute x = 2 into one equation to find y
A1 y = 5, intersection (2, 5) cao
Answer: (2, 5)

Question 6

(a) M1 substitute (6, -2) into y = ax + 4: -2 = 6a + 4 and form 6a = -6
(a) A1 a = -1 cao
(a) Answer: a = -1
(b) M1 use perpendicular gradient product = -1: gradient of L is -1 so perpendicular gradient = 1
(b) M1 use point (6, -2) to form equation y = 1x + c and solve for c: -2 = 6 + c
(b) A1 y = x - 8 cao
(b) Answer: y = x - 8

Question 7

(a) B1 booking fee identified as C when h = 0 is 18 pounds
(a) M1 find gradient (58 - 18) / (5 - 0) = 40 / 5 = 8
(a) A1 hourly rate 8 pounds per hour cao
(a) Answer: Booking fee = 18 pounds; hourly rate = 8 pounds/hour
(b) M1 use C = mh + c with m = 8 and c = 18
(b) A1 C = 8h + 18 cao
(b) A1 units consistent: C in pounds and h in hours (communication) oe
(b) Answer: C = 8h + 18

Question 8

(a) M1 recognise rearrangement or set up to eliminate m: write y - mx = 3 - 2m
(a) M1 collect m terms: y - mx = 3 - 2m => bring m terms together: y - 3 = m(x - 2) or equivalent
(a) M1 argument that for a fixed point (x0, y0) independent of m, coefficient of m must be zero: require x - 2 = 0 so x = 2
(a) M1 substitute x = 2 into original to find y: y = m*2 + (3 - 2m) => simplify
(a) A1 y = 2m + 3 - 2m = 3, so fixed y = 3 cao
(a) A1 conclude fixed point is (2, 3) cao
(a) M1 give a clear statement that for any m the point (2, 3) satisfies y = mx + 3 - 2m
(a) A1 (2, 3) is on every line in the family cao
(a) Answer: (2, 3)

Question 9

(a) B1 350 is speed in metres per minute oe
(a) B1 120 is initial distance from the start in metres at t = 0 (a head start) oe
(a) Answer: 350 metres per minute; 120 metres initial head start
(b) M1 set 350t + 120 = 450(t - 6), correctly using the second cyclist's delayed start
(b) M1 expand and rearrange to 2820 = 100t
(b) A1 t = 28.2 minutes cao
(b) M1 check second cyclist has actually started: confirm t = 28.2 is greater than 6, so the model d = 450(t - 6) applies
(b) A1 state that catching occurs at t = 28.2 minutes since original start cao
(b) B1 units stated: minutes oe
(b) M1 sanity check: distances equal: 350*28.2 + 120 = 9990 and 450*(28.2 - 6) = 9990
(b) A1 final value 28.2 minutes cao
(b) Answer: 28.2 minutes

Question 10

(a) M1 isolate term -2y = -4x -6 or equivalent
(a) M1 divide by -2 to get y = 2x + 3
(a) A1 y = 2x + 3 cao
(a) Answer: y = 2x + 3
(b) M1 understand translation shifts every point: use general point (x, y) on L1 and map to (x + 3, y + 2) or shift intercepts
(b) M1 substitute x' = x + 3, y' = y + 2 into y = 2x + 3 or transform intercept: y' - 2 = 2(x' - 3) + 3
(b) M1 simplify to find equation in y' and x': y' = 2x' - 1 or equivalent correct algebra
(b) B1 sketch shows two parallel lines with correct relative shift (L3 above and right of L1) oe
(b) A1 final equation L3: y = 2x - 1 cao
(b) Answer: y = 2x - 1

Drawing Linear Graphs: Higher Tier Practice - Worksheets, Questions and Revision