New Effective Learning Mathematics Chapter Test | S1B Suggested Solutions
Chapter 9 Introduction to the Rectangular Coordinate System
Ans: C D B
Explanation:
1. Distance between (9, –3) and P(4, –3) = (9 – 4) units
= 5 units
Distance between (–1, –3) and P(4, –3) = [4 – (–1)] units
= 5 units
Distance between (4, –2) and P(4, –3) = [–2 – (–3)] units
= 1 unit
Distance between (4, 2) and P(4, –3) = [2 – (–3)] units
= 5 units
The answer is C. 1A
(1)
2. Coordinates of the image of A = (a 3, 5)
Coordinates of the image of B = (b + 4, 5)
Distance between the images of A and B = [(b + 4) (a 3)] units
= (b + 4 a + 3) units
= (b a + 7) units
The answer is D. 1A
(1)
3. Perpendicular distance from P to L = (3 – k) units
Perpendicular distance from Q to L = [k – (–9)] units
= (k + 9) units
3 k k 9
2k 6
k 3
The answer is B. 1A
(1)
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New Effective Learning Mathematics Chapter Test | S1B Suggested Solutions
Chapter 9 Introduction to the Rectangular Coordinate System
4. (a) Let BC = x units, then AB = (x – 1) units and CD = (x + 2) units.
x ( x 1) ( x 2) 28 1M
3x 1 28
3x 27
x9
The length of BC is 9 units. 1A
(2)
(b) (i) AB = (9 – 1) units
= 8 units
CD = (9 + 2) units
= 11 units
Coordinates of B (3 8, 6) 1M
(5, 6)
Coordinates of C (5, 6 9) 1M
(5, 3)
Coordinates of D (5 11, 3)
(6, 3) 1A
(ii) A is rotated anticlockwise about the origin through 90 to D. 1A
(or A is rotated clockwise about the origin through 270 to D.)
(4)
5. (a)
1A + 1A
(2)
(b) B lies on the side of ABC. 1A
(1)
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New Effective Learning Mathematics Chapter Test | S1B Suggested Solutions
Chapter 9 Introduction to the Rectangular Coordinate System
6. (a) Draw trapezium ADEC such that AD and CE are horizontal line segments,
DBE is a vertical line segment.
CE = [7 – (–3)] units
= 10 units
AD = [7 – (–5)] units
= 12 units
DE = [5 – (–1)] units
= 6 units
1
Area of trapezium ADEC = (10 12) 6 sq. units
2
= 66 sq. units
BD = [4 – (–1)] units 1A
= 5 units
1
Area of ADB = 12 5 sq. units
2
= 30 sq. units
BE = (5 – 4) unit
= 1 unit
1
Area of BEC 10 1 sq. units
2
5 sq. units
Area of ABC = Area of trapezium ADEC Area of ADB Area of BEC
= (66 – 30 – 5) sq. units 1M
= 31 sq. units 1A
(3)
(b) Let h units be the perpendicular distance from C to AB.
Area of ABC = 31 sq. units
1
13 h 31 1M
2
62
h
13
62
The perpendicular distance from C to AB is units. 1A
13
(2)
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New Effective Learning Mathematics Chapter Test | S1B Suggested Solutions
Chapter 9 Introduction to the Rectangular Coordinate System
7. (a) AB // DC and the x-coordinates of A and B are the same.
x-coordinate of D = x-coordinate of C
= –1 1A
(1)
(b) Height of ABCD = [–1 – (–5)] units
= 4 units
AB = (8 – 1) units
= 7 units
Let DC = b units.
Area of ABCD = 20 sq. units
1
(7 + b) 4 = 20 1M
2
7 + b = 10
b =3
DC = 3 units
y-coordinate of C y-coordinate of D = 3
y-coordinate of D = 5 3 1M
=2 1A
(3)
(c) The coordinates of D are (–1, 2).
After the reflection of D, coordinates of D = (–1, 2) 1M
After the translation of D, coordinates of D = (1 + 5, –2)
= (4, –2) 1A
(2)
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New Effective Learning Mathematics Chapter Test | S1B Suggested Solutions
Chapter 9 Introduction to the Rectangular Coordinate System
8. (a) Coordinates of Q = (5, (k))
= (5, k) 1A
Coordinates of R = (5 + 35, k)
= (30, k) 1A
(2)
(b) PQ = k – (–k)
= 2k
QR = 35
Perimeter of PQR = PQ + QR + PR
84 = 2k + 35 + 37 1M
2k = 12
k =6 1A
(2)
1
(c) Area of PQR = 35 2(6) sq. units 1M
2
= 210 sq. units 1A
(2)
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