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13+ · Mathematics

CE.M10 Coordinate Geometry and Real-Life Graphs

ISEB CE CE.M10 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The grid shows two points, A(1, -3) and C(-2, 1).

(a)Write down the coordinates of the point that is 4 units to the right of A, along a horizontal line.(1)

(b)Write down the coordinates of the point that is 5 units above C, along a vertical line.(1)

(Total for Question 1 is 2 marks)

A is the point (2, 3) and B is the point (8, 11).

(Total for Question 2 is 2 marks)

A straight line passes through the points (1, 4) and (5, 12).

(Total for Question 3 is 2 marks)

PQRS is a rectangle. P is the point (1, 1), Q is the point (6, 1) and R is the point (6, 4), as shown on the grid.

(Total for Question 4 is 3 marks)

E is the point (-1, 2) and F is the point (5, 10), as shown on the grid.

(Total for Question 5 is 3 marks)

The graph shows a straight line.

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

(b)Write down the equation of the line in the form y = mx + c.(2)

(Total for Question 6 is 4 marks)

The graph converts British pounds (GBP) to euros (EUR).

(a)Use the graph to find the rate of conversion, in euros per pound, and use it to work out how many euros GBP 35 would convert to.(2)

(b)Anna has 90 euros. Work out how many pounds this is.(2)

(Total for Question 7 is 4 marks)

The graph shows a cyclist's distance from home during a 45-minute journey.

(a)Work out the cyclist's speed, in km/h, for the first 15 minutes of the journey.(3)

(b)State what is happening between t = 15 minutes and t = 25 minutes, giving a reason from the graph.(1)

(c)Work out the average speed, in km/h, for the final 20 minutes of the journey.(3)

(Total for Question 8 is 7 marks)

The graph is used to convert between degrees Celsius (C) and degrees Fahrenheit (F).

(a)Find the gradient of the conversion graph, taking C as the horizontal axis and F as the vertical axis.(2)

(b)Write down the equation connecting F and C in the form F = mC + c, and use it to convert 35 degrees Celsius to Fahrenheit.(3)

(Total for Question 9 is 5 marks)

Triangle XYZ has vertices X(0, 0), Y(6, 0) and Z(3, 4), as shown on the grid.

(Total for Question 10 is 4 marks)

A line L1 has equation y = 3x - 2. Line L2 is parallel to L1 and passes through the point (2, 9).

(Total for Question 11 is 3 marks)

A line has equation y = (2/5)x + 1.

(a)Write down the gradient of a line perpendicular to this line.(2)

(b)This perpendicular line passes through the point (10, -3). Find its equation.(2)

(Total for Question 12 is 4 marks)

A(1, 2), B(4, 8) and C(6, 12) are three points.

(Total for Question 13 is 3 marks)

Water is poured into an empty tank at a steady rate. The graph shows the volume of water in the tank for the first 10 minutes.

(a)Find the rate at which water is poured into the tank, in litres per minute.(2)

(b)The tank has a total capacity of 65 litres. Work out how long it takes, in total, to fill the tank completely.(2)

(Total for Question 14 is 4 marks)

Tariff A charges a fixed GBP 10 per month plus GBP 0.05 per minute of calls. Tariff B charges no fixed fee but GBP 0.15 per minute of calls. The graph shows both tariffs (Tariff A solid, Tariff B dashed) for m minutes of calls.

(a)Write down expressions for the monthly cost, in pounds, of Tariff A and of Tariff B, for m minutes of calls in a month.(2)

(b)Find the number of minutes of calls for which the two tariffs cost the same amount, and state which tariff is cheaper for a customer who uses 200 minutes in a month.(3)

(Total for Question 15 is 5 marks)

A car hire company charges a fixed booking fee plus a charge per day. The graph shows the total cost for different numbers of days hired.

(a)Use the graph to find the daily hire charge and the fixed booking fee.(3)

(b)Use your answers to part (a) to find the total cost of hiring the car for 12 days.(2)

(Total for Question 16 is 5 marks)

The midpoint of the line segment PQ is M(4, -1). P has coordinates (-2, 3).

(Total for Question 17 is 3 marks)

A quadrilateral has vertices A(0, 1), B(4, 3), C(6, -1) and D(2, -3), as shown on the grid.

(a)Calculate the length of side AB, giving your answer correct to 1 decimal place.(2)

(b)Show that AB is parallel to DC.(2)

(c)Given that AB is also equal in length to DC, state what type of quadrilateral ABCD is, giving a reason.(2)

(Total for Question 18 is 6 marks)

The line segment joining A(-3, -1) and B(9, 7), shown on the grid, is divided by the point R such that AR:RB = 1:3.

(a)Find the coordinates of R.(3)

(b)Verify your answer to part (a) by checking that R is three times as close to A as it is to B.(2)

(Total for Question 19 is 5 marks)

Mark scheme · CE.M10 Coordinate Geometry and Real-Life Graphs

Question 1

(a) B1 (5, -3) cao
(a) Answer: (5, -3)
(b) B1 (-2, 6) cao
(b) Answer: (-2, 6)

Question 2

M1 ((2+8)/2, (3+11)/2) oe seen
A1 (5, 7) cao
Answer: (5, 7)

Question 3

M1 (12-4)/(5-1) oe seen
A1 2 cao
Answer: gradient = 2

Question 4

M1 recognises that PS must be parallel and equal in length to QR (or equivalent vector reasoning)
M1 S = P + (R - Q), i.e. S = (1, 1) + (0, 3), or equivalent correct method
A1 (1, 4) cao
Answer: S = (1, 4)

Question 5

M1 horizontal difference = 6 and vertical difference = 8 identified (or squared terms shown)
M1 √6² + 8² seen
A1 10 cao
Answer: EF = 10

Question 6

(a) M1 uses two points on the line, e.g. (0,3) and (2,7), (7-3)/(2-0) oe
(a) A1 2 cao
(a) Answer: gradient = 2
(b) M1 identifies c = 3 from the graph (y-intercept at x = 0)
(b) A1 y = 2x + 3 oe cao, ft from part a
(b) Answer: y = 2x + 3

Question 7

(a) M1 rate = 24/20 = 1.2 (or 60/50 = 1.2) euros per pound seen, then 35 x 1.2
(a) A1 42 (euros) cao
(a) Answer: 42 euros
(b) M1 90 / 1.2 oe, ft from part a
(b) A1 GBP 75 cao
(b) Answer: GBP 75

Question 8

(a) M1 15 minutes = 0.25 hours (or equivalent, e.g. 1/4)
(a) M1 6 / 0.25 oe, dependent on previous method
(a) A1 24 km/h cao
(a) Answer: 24 km/h
(b) B1 the cyclist is stationary (resting), since the distance from home stays at 6 km oe
(b) Answer: The cyclist is stationary (resting) - the distance from home stays constant at 6 km.
(c) M1 distance covered in final section = 15 - 6 = 9 km
(c) M1 20 minutes = 1/3 hour, then 9 / (1/3) oe, dependent on previous method
(c) A1 27 km/h cao (awrt 27)
(c) Answer: 27 km/h

Question 9

(a) M1 (50-32)/(10-0) oe seen
(a) A1 1.8 cao
(a) Answer: 1.8
(b) M1 F = 1.8C + 32 oe, using intercept 32 from the graph and gradient from part a
(b) M1 1.8 x 35 + 32, dependent on previous method
(b) A1 95 (degrees Fahrenheit) cao
(b) Answer: F = 1.8C + 32; 35 degrees C = 95 degrees F

Question 10

M1 XY = 6 (horizontal distance)
M1 XZ = √3² + 4² = 5
M1 YZ = √3² + 4² = 5
A1 perimeter = 16 cao
Answer: perimeter = 16

Question 11

M1 gradient of L2 = 3 (parallel lines have equal gradient)
M1 9 = 3(2) + c, dependent on previous method
A1 y = 3x + 3 cao
Answer: y = 3x + 3

Question 12

(a) M1 uses gradient x gradient = -1 (or equivalent perpendicular gradient rule)
(a) A1 -5/2 oe cao
(a) Answer: -5/2
(b) M1 -3 = (-5/2)(10) + c, ft from part a
(b) A1 y = -2.5x + 22 oe cao
(b) Answer: y = -2.5x + 22 (or y = -5/2 x + 22)

Question 13

M1 gradient of AB = (8-2)/(4-1) = 2
M1 gradient of BC = (12-8)/(6-4) = 2
A1 gradients are equal and B is common to both lines, so A, B and C are collinear (cso)
Answer: A, B and C are collinear, since gradient AB = gradient BC = 2

Question 14

(a) M1 (50-20)/(10-4) oe seen
(a) A1 5 litres per minute cao
(a) Answer: 5 litres per minute
(b) M1 65/5 oe, ft from part a (uses V = 5t, since the tank starts empty)
(b) A1 13 minutes cao
(b) Answer: 13 minutes

Question 15

(a) B1 Tariff A: C = 10 + 0.05m oe
(a) B1 Tariff B: C = 0.15m oe
(a) Answer: Tariff A: C = 10 + 0.05m; Tariff B: C = 0.15m
(b) M1 10 + 0.05m = 0.15m oe, ft from part a
(b) M1 solves to m = 100, dependent on previous method
(b) A1 Tariff A is cheaper at 200 minutes (Tariff A = GBP 20, Tariff B = GBP 30) cao
(b) Answer: m = 100 minutes; at 200 minutes, Tariff A is cheaper (GBP 20 compared with GBP 30)

Question 16

(a) M1 (130-70)/(5-2) = 20 (daily charge) oe
(a) M1 70 = 20(2) + c, dependent on previous method
(a) A1 fixed fee = GBP 30 cao
(a) Answer: daily charge = GBP 20; fixed fee = GBP 30
(b) M1 20(12) + 30 oe, ft from part a
(b) A1 GBP 270 cao
(b) Answer: GBP 270

Question 17

M1 Qx = 2(4) - (-2), dependent on correct use of the midpoint formula
M1 Qy = 2(-1) - 3
A1 (10, -5) cao
Answer: Q = (10, -5)

Question 18

(a) M1 √4² + 2² seen
(a) A1 4.5 cao (awrt 4.5)
(a) Answer: AB = 4.5 (√20 = 4.47 to 3 sf)
(b) M1 gradient AB = (3-1)/(4-0) = 0.5
(b) A1 gradient DC = (-1-(-3))/(6-2) = 0.5; equal gradients, so AB is parallel to DC (cso)
(b) Answer: gradient AB = gradient DC = 0.5, so AB is parallel to DC
(c) B1 parallelogram
(c) B1 reason: one pair of opposite sides, AB and DC, are equal in length and parallel oe
(c) Answer: ABCD is a parallelogram, because AB and DC are equal in length and parallel

Question 19

(a) M1 B - A = (12, 8) found
(a) M1 R = A + (1/4)(12, 8), dependent on previous method
(a) A1 (0, 1) cao
(a) Answer: R = (0, 1)
(b) M1 AR = √3²+2² = √13; RB = √9²+6² = √117, ft from part a
(b) A1 RB = 3 x AR (√117 = 3sqrt(13)), confirming the ratio 1:3 (cso)
(b) Answer: AR = √13, RB = 3sqrt(13), so RB = 3 x AR, confirming AR:RB = 1:3

Coordinate Geometry and Real-Life Graphs - Worksheets, Questions and Revision