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CHAPTER Straight Lines Mathematics Form 3 Chapter 9 Straight Lines 9. (a) LHS: 3(2) = 6

y
x
9
(b) = – + 1

5
10
RHS: 2(0) + 6 = 6

Conclusion: Yes
5x
Garis Lurus
y = – + 5
10
1 (b) LHS: 4(–2) = –8
y = – x + 5
2 RHS: 3(–1) + 1 = –2
Conclusion: No
1. (a) m = undefined –7 –7 y x
(c) Gradient m = = (c) – = – + 1
(b) m < 0 –7 11 – 5 6 4 3 (c) LHS: 0
(c) m = 0 y = 6 x + c, y = 4x – 4 RHS: 5(–3) – 1 = –16
3
substitute (5, 0) Conclusion: No
y
2. (a) y = –5x + 2 0 = –7 (5) + c (d) – = 4x + 1
9
6
(b) y = 5 35 9 y = –4x – 9 10. (a) L 1 : 3y = –6x + 7
7
(c) y = 3 c = 6 y = –2x + , gradient m 1 = –2
1 3
7
(d) y = x y = – x + 35
1
2 6 6 7. (a) y = x – 9 L 2 : –4y = 8x – 5
3 2 6 5
(e) y = x + y = –2x + , gradient m 2 = –2
4 5 6y = x – 54 4
5. (a) 3y = –5x + 2
–5 2 x – 6y – 54 = 0 m 1 = m 2 , therefore L 1 is parallel to L 2 .
3. (a) m = –5, c = –6 y = 3 x + 3
1
(b) m = , c = 7 x y 1
2 (b) 6y = 2x +3 (b) 2 + = 1 (b) L 1 : 2y + 1 = – x + 6
8
(c) m = 0, c = 8 2 3 x y 2

y = x +  + = 1 × 8 1 5 1
6 6 2 8 y = – x + , gradient m 1 = –
4
4
2
3 – 5 1 1 1 4x + y = 8
4. (a) Gradient m = –4 – 0 = 2 y = x + 2 L 2 : 3y = 8x – 3
3
1 4x + y – 8 = 0
y = x + c, 8 8
2 y = x – 1, gradient m 2 =
(c) 4y = –x – 1 3 3
substitute (0, 5) x 1
y = – – 8. (a) 5y = 2x + 10
5 = 0 + c 4 4 m 1 ≠ m 2 , therefore L 1 is not parallel to
–2x + 5y = 10
c = 5 2x 5y 10 L 2 .
1 (d) –8y = –4x +12 – + =
y = x + 5 10 10 10
2 –4 12 x
y
y = x – – + = 1 (c) L 1 : 3y + 2x = –6
–8 8 5 2
2
2
–2 – 1 1 3 y = – x – 2, gradient m 1 = –
(b) Gradient m = y = x –
10 – 9 2 2 (b) 6y – 2x – 8 = 0 3 3
–3 (–2x + 6y = 8) ÷ 8 L 2 : 6 – 5y = 8x – 1
=
1 y x –x 3y 8 7 8
= –3 6. (a) = – + 1 4 + 4 = 1 y = – x + , gradient m 2 = – 5
8
4
5
5
8
y = –3x + c, y = – x + 8 1
substitute (9, 1) 4 (c) y = x – 6 m 1 ≠ m 2 , therefore L 1 is not parallel to
2
y = –2x + 8

1 = –3(9) + c  1 x – y = 6 ÷ 6 L 2 .
c = 28 2
1
y
y = –3x + 28 x – = 1
12 6
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BOOKLET ANS MATH F3.indd 25 03/01/2020 10:21 AM