More IB linear Questions 1.
Consider the line l: 2x + y + 4 = 0.
(a)
Write down the gradient of l.
(b)
Find the gradient of a line perpendicular to l.
(c)
Find the equation of a line perpendicular to l, passing through the point (5, 3). Give your answer in
the form ax + by + d = 0, where a, b, d ∈ .
(Total 6 marks)
2.
The following diagram shows the points P, Q and M. M is the midpoint of [PQ].
y
P(0, 2)
M
0
Q(2, 0)
x
(a)
Write down the equation of the line (PQ).
(b)
Write down the equation of the line through M which is perpendicular to the line (PQ).
(Total 4 marks)
3.
The diagram below shows the line with equation 3x + 2y = 18. The points A and B are the y and xintercepts respectively. M is the midpoint of [AB].
y
A
3x + 2y =18
M
Diagram not to scale
B
O
x
Find the coordinates of
(a)
the point A;
(b)
the point B;
(c)
the point M.
(Total 8 marks)
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)
5.
At Jumbo’s Burger Bar, Jumbo burgers cost £J each and regular cokes cost £C each. Two Jumbo burgers
and three regular cokes cost £5.95.
(a)
Write an equation to show this.
(b)
If one Jumbo Burger costs £2.15, what is the cost, in pence, of one regular coke?
(Total 4 marks)
6.
The conversion formula for temperature from the Fahrenheit (F) to the Celsius (C) scale is given by C =
5( F – 32)
.
9
(a)
What is the temperature in degrees Celsius when it is 50° Fahrenheit?
There is another temperature scale called the Kelvin (K) scale.
The temperature in degrees Kelvin is given by K = C + 273.
(b)
What is the temperature in Fahrenheit when it is zero degrees on the Kelvin scale?
(Total 8 marks)
7.
The line L1 shown on the set of axes below has equation 3x + 4y = 24. L1 cuts the x-axis at A and cuts the
y-axis at B.
Diagram not drawn to scale
y
L1
B
L2
M
O
A
x
C
(a)
Write down the coordinates of A and B.
(2)
M is the midpoint of the line segment [AB].
(b)
Write down the coordinates of M.
(2)
The line L2 passes through the point M and the point C (0, –2).
(c)
Write down the equation of L2.
(2)
(d)
Find the length of
(i)
MC;
(2)
(ii)
AC.
8.
The following table gives the postage rates for sending letters from the Netherlands. All prices are given
in Euros (€).
Destination
Weight not more than 20 g Each additional 20 g or part of 20 g
Within the Netherlands
(zone 1)
€0.40
€0.35
Other destinations
within Europe (zone 2)
€0.55
€0.50
Outside Europe
(zone 3)
€0.80
€0.70
(a)
Write down the cost of sending a letter weighing 15 g from the Netherlands to a destination within
the Netherlands (zone 1).
(b)
Find the cost of sending a letter weighing 35 g from the Netherlands to a destination in France
(zone 2).
(c)
Find the cost of sending a letter weighing 50 g from the Netherlands to a destination in the USA
(zone 3).
(Total 8 marks)
9.
A is the point (2, 3), and B is the point (4, 9).
(a)
Find the gradient of the line segment [AB].
(b)
Find the gradient of a line perpendicular to the line segment [AB].
(c)
The line 2x + by – 12 = 0 is perpendicular to the line segment [AB]. What is the value of b?
(Total 4 marks)
Answers 1.
(a)
–2
(A1) (C1)
(b)
gradient =
(c)
Using y = mx + c with their m to find c
1
c=
2
1
1
y= x+
2
2
x – 2y + l = 0
1
2
(A1)
(C1)
(M1)
(A1)
(A1)
(A1)
(C4)
[6]
2.
(a)
y = –x + 2 or x + y = 2 or x + y – 2 = 0
(A1)
(b)
⎛ 0 + 2 2 + 0 ⎞
Midpoint M: ⎜
,
⎟ = (1, 1)
2 ⎠
⎝ 2
(M1)
gradient = 1
(A1)
1 = 1(1) + (b) ⇒ b = 0
y=x
(A1)
[4]
3.
(a)
(b)
3x + 2y = 18
2x = 18
y=9
therefore A = (0,9)
3x = 18
x=6
B = (6,0)
(A1)
(A1) (C2)
(A1)
(A1)
(C2)
Note: Award (A0), (A1) (ft) for A = 9, B = 6.
Award (A0), (A2) (ft) for A = (0, 6) and B = (9, 0).
(c)
(0, 9) (6, 0)
⎛ 0 + 6 9 + 0 ⎞
midpoint = ⎜
,
⎟
2 ⎠
⎝ 2
= (3, 4.5)
(A1)
(A1)
(A1)(A1)
(C2)(C2)
[8]
4.
(a)
(b)
0 − (−2)
Gradient of l 2 =
5−0
2
= 5
−5
l =
Gradient of 1
2
−5
x+7
2
2y = –5x + 14
5x + 2y – 14 = 0
(M1)
(A1)
(A1) (C3)
y=
(A1)(A1)
(A1)(A1)(A1)
(C5)
[8]
5.
(a)
2J + 3C = 5.95
(b)
2×2.15 + 3C = 5.95
3C = 1.65
C = 0.55
55 (pence) or £0.55
(A2) (C2)
(M1)
(M1)
(A1)
(C2)
[4]
6.
(a)
(b)
5(50 − 32)
9
= 10°C.
C=
(M1)
(A2) (C3)
Put C = –273
5( F − 32)
so – 273 =
9
Hence 9 × –273 = 5(F – 32)
F = –491.4 + 32 = –459.4 (accept –459).
(A1)
(M1)
(M1)
(M1)
(C5)
(A1)
Note: (M1) is for adding 32, even if the other number is incorrect.
[8]
7.
(a)
A; y = 0, 3x = 24 ⇒ x = 8
A(8, 0)
B;
(b)
x = 0, 4y = 24 ⇒ y = 6
B(0, 6)
M; xm =
M(4, 3)
8+0
0+6
= 4, ym =
=3
2
2
(A1)
(A1)
2
(A1)
2
(A1)
(c)
(d)
3 − −2 5
=
L2: gradient =
4−0 4
5
y = 4 x – 2 (or equivalent)
(i)
(A1)
(4 − 0) 2 + (3 − (−2)) 2
= 41
= 6.40
(M1)
(A1)
A(8, 0), C(0, –2)
AC =
8 2 + (−2) 2
(M1)
= 68
= 8.25
8.
2
M(4, 3), C(0, –2)
MC =
(ii)
(A1)
(a)
0.40
(b)
0.55 + 0.50 = 1.05
(A1)
4
(A2) (C2)
(A1)(
(C3)
A1)(A1)
Note: Award (A1) for 0.55, (A1) for 0.50, (A1) for correct total of
amounts given.
(c)
0.80 + 1.40 = 2.20
(A1)(
(C3)
A1)(A1)
Note: Award (A1) for 0.80, (A1) for 1.40, (A1) for correct total of
amounts given.
[8]
9.
(a)
For the line (AB), m =
6
2
=3
9−3
4−2
=
(A1)
1
3
(b)
m=–
(c)
2x + by – 12 = 0
2
12
x+
b
b
1
2
Therefore, − = −
3
b
(A1)
y= −
6=b
(M1)
(A1)
[4]