IB Review Questions – Linear Functions
1.
The diagram below shows the line PQ, whose equation is x + 2y = 12. The line intercepts the axes at P and
Q respectively.
diagram not to scale
(a)
Find the coordinates of P and of Q.
(3)
(b)
A second line with equation x – y = 3 intersects the line PQ at the point A. Find the coordinates of A.
(3)
(Total 6 marks)
IB Questionbank Mathematical Studies 3rd edition
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2.
A student has drawn the two straight line graphs L1 and L2 and marked in the angle between them as a right
angle, as shown below. The student has drawn one of the lines incorrectly.
y
3
L2
90°
2
1
–4
–3
–2
–1
0
1
2
3
4
x
–1
L1
Consider L1 with equation y = 2x + 2 and L2 with equation y = –
1
x + 1.
4
(a)
Write down the gradients of L1 and L2 using the given equations.
(b)
Which of the two lines has the student drawn incorrectly?
(c)
How can you tell from the answer to part (a) that the angle between L1 and L2 should not be 90°?
(d)
Draw the correct version of the incorrectly drawn line on the diagram.
(Total 8 marks)
IB Questionbank Mathematical Studies 3rd edition
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3.
The four diagrams below show the graphs of four different straight lines, all drawn to the same scale. Each
diagram is numbered and c is a positive constant.
y
y
c
c
Number 1
Number 3
x
0
x
0
y
c
y
c
Number 2
Number 4
x
0
0
x
In the table below, write the number of the diagram whose straight line corresponds to the equation in the
table.
Equation
Diagram number
y=c
y=–x+c
y=3x+c
y=
1
x+c
3
(Total 8 marks)
IB Questionbank Mathematical Studies 3rd edition
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4.
The following diagram shows the lines l1 and l2, which are perpendicular to each other.
Diagram not to scale
y
(0, 7)
l2
(0, –2)
(5, 0)
x
l1
(a)
Calculate the gradient of line l1.
(b)
Write the equation of line l1 in the form ax + by + d = 0 where a, b and d are integers, and a > 0.
(Total 8 marks)
IB Questionbank Mathematical Studies 3rd edition
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5.
The costs charged by two taxi services are represented by the two parallel lines on the following graph. The
Speedy Taxi Service charges $1.80, plus 10 cents for each kilometre.
c
2.60
Speedy Taxi Service
2.40
2.20
2.00
cost ($)
1.80
1.60
Economic Taxi Service
1.40
1.20
1.00
0
2
4
6
8
distance (km)
k
(a)
Write an equation for the cost, c, in $, of using the Economic Taxi Service for any number of
kilometres, k.
(b)
Bruce uses the Economic Taxi Service.
(i)
How much will he pay for travelling 7 km?
(ii)
How far can he travel for $2.40?
(Total 4 marks)
IB Questionbank Mathematical Studies 3rd edition
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SOLUTIONS
1.
(a)
c – 0.10k + 1.40
(b)
(i)
(ii)
c = 0.10(7) + 1.40 (allow follow through from part (a))
= 0.70 + 1.40
= $2.10
2.40 = 0.10k + 1.40 (allow follow through from part (a))
1.00 = 0.10k
10 = k
10 km
(A1)
(A1)
(M1)
(A1)
[4]
2.
(a)
0 + 2y = 12 or x + 2(0) = 12
(M1)
P(0, 6)
(accept x = 0, y = 6)
(A1)
Q(12,0)
(accept x = 12, y = 0)
(A1) (C3)
Notes: Award (M1) for setting either value to zero.
Missing coordinate brackets receive (A0) the first time this occurs. Award
(A0)(A1)(ft) for P(0,12) and Q(6, 0).
(b)
x + 2(x – 3) = 12
(M1)
(6, 3)
(accept x = 6, y = 3)
(A1)(A1) (C3)
Note: (A1) for each correct coordinate.
Missing coordinate brackets receive (A0)(A1) if this is the first time it
occurs.
[6]
3.
1
.
(A1)(A1) (C2)
4
Note: Award (A0)(A1)ft if the order of the gradients is reversed or both
signs are wrong or both are reciprocals of the correct answer.
(a)
L1 has gradient 2 and L2 has gradient 
(b)
L2 is drawn incorrectly.
(c)
The product of the gradients is 2 × 
(A2) (C2)
1
1
(M1)(A1) (C2)
   –1.
4
2
Note: Award (M1) for looking at product of gradients,
(A1) for comparing something to –1.
IB Questionbank Mathematical Studies 3rd edition
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(d)
The drawing should show a straight line passing through
x and y intercepts at (4, 0) and (0, 1) respectively.
(A1)(A1) (C2)
Note: Award (A1) for each intercept. If these are wrong but gradient is
1
then (A1). If correct line is very poorly drawn then (A1).

4
[8]
4.
Equation
Diagram number
y=c
2
(A2)
y = –x + c
3
(A2)
y = 3x + c
4
(A2)
1
(A2) (C8)
1
x+c
3
y=
[8]
5.
(a)
Gradient of l 2 
=
2
5
Gradient of l1 
(b)
0  (2)
50
(M1)
(A1)
5
2
5
x+7
2
2y = –5x + 14
5x + 2y – 14 = 0
y=
(A1) (C3)
(A1)(A1)
(A1)(A1)(A1) (C5)
[8]
IB Questionbank Mathematical Studies 3rd edition
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