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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines

PT3 Standard Practice 9 Section B (b) Equation of line PQ 2. (a) (i) k located at line y = 3
y = – 5 x + 5 ..............(1) Therefore k(a, 3)
Section A 1. Diagram Equation 3 y = 2x – 4
Rajah Persamaan Equation of line RS
6 – (–4) 3 = 2(a) – 4
1. Gradient of KL = (a) y y = x + 3 ....................(2)
8 – (–2) a = 3.5
= 1 K 2 y = – + 2 (1) = (2), ∴ k(3.5, 3)
2

Equation of KL x 5  – 5 x + 5 = x + 3 × 3
y = x + c O 5 L 3 (ii) x = –4
at L(8, 6) (b) –5x + 15 = 3x + 9 (iii) m ML = m KN
6 = 8 + c y x = 3 = 2
c = –2 K(8, 4) 4
1 3 y = 2x + c
∴ y = x – 2 y = x Substitute x = into (2)
x 2 4 Substitute x = –4, y = 0,
Answer: D L O y = 3 + 3
4 0 = 2(–4) + c
2. 4x + 3y = 15 y = 15 c = 8
3y = –4x + 15 2. Equation Point 4 ∴ y = 2x + 8
4 Persamaan Titik 3 15
y = – x + 5 ∴ k  ,  (iv) Coordinates P = (p, 0)
3 4 4
(a) y = 3x – 7 (6, 11) ✓ Substitute x = p, y = 0 into y = 2x – 4,
4
∴ m = – , c = 5 (c) Substitute x = 2, y = 8, 0 = 2p – 4
3
(b) –5x + y = 2 (–3, –13) ✓ y = k x + 5m p = 2
Answer: B 3 Length of MP = 2 – (–4)
1
3. x-intercept, y = 0 (c) 3y = x + 8 (2, –3) ✗ 8 = 2k + 5m = 6 units
2 3
2x – 3y = 6 24 = 2k + 15m ...................(1) Area of KLMP = (3 × 6)
1
2x – 3(0) = 6 (d) 1 y + x = 1 (–10, 6) ✗ Substitute x = –5, y = –13 = 18 units 2
2x = 6 3 5 k
x = 3 y = x + 5m (b) 2x + 3y – 4 = 0
Section C 3 3y = –2x + 4
∴ x-intercept = 3 –13 = –5k + 5m
2
1. (a) (i) Rhombus has equal length of sides. 3 y = – x + 4
Answer: B 3 3
OK = KL = LM = MO –39 = –5k + 15m ...............(2) 2
2
4. m KL = –1, c = –5 OK = (–5) + 12 2 (1) – (2) gradient, m = – 3
= 13 units 24 – (–39) = 2k – (–5k)
∴ equation of the straight line KL is (c) 5x + 2y = 8
y = –x – 5 ∴ OM = 13 units 63 = 7k
M(–13, 0) k = 9 Divide both sides by 8,
Answer: C Substitute k = 9 into (1) 5x 2y 8
(ii) m KO = 12 – 0 8 + 8 = 8
2 –5 – (0) 24 = 2(9) + 15m
5. y – x = 5 x y
3 = – 12 15m = 6 8 + = 1
4
3y – 2x = 15 5 2
2x – 3y + 15 = 0 y = mx + c m = 5 5
12 Therefore, x-intercept = 8
Answer: C y = – x + c ∴ m = 2 , k = 9 5
5 5 y-intercept = 4
Substitute x = 0, y = 0,
c = 0
Equation of LM

y = – 12 x
5
(iii) y = 12
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BOOKLET ANS MATH F3.indd 28 03/01/2020 10:21 AM