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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
11. (a) 2y = 6 – 3x 12. (a) y = 2x – 4 (b) y = –x – 2 From the graph,
–3
y = x +3 x y x y Point of intersection = (2, 0)
2 y
3 0 –4 0 –2
m = –
2 2 0 –2 0
new equation y = –x + 2 4
3 y = 6 – 2x y = 2x – 2
y = – x + c
2 x y x y 2
1
2
pada titik (5, 0) 0 6 0 –2 y = –x – 1
3
0 = – (5) + c 3 0
2 1 0 –4 –2 0 2 4 x
15
c = From the graph,
2 From the graph, –2
Point of intersection
3 15
Maka, y = – x + Point of intersection
2 2 = (2.5, 1)
= (0, –2)
y = 6 – 2x
(b) 6x – 3y = 8 y y
6 8 13. (a) y = 2x – 1 ................(1)
y = x – 6 y = 3 – x ..................(2)
3 3 4 y = 2x – 2
8
= 2x – (1) – (2)
3 4 y = 2x – 4 3x – 4 = 0
m = 2 y = –x – 2 2 1
x = 1
new equation 2 3
1
y = 2x + c –4 –2 0 2 4 x Substitute x = 1 into (2)
(2.5, 1)
3
at the point (8, –1) x 1 2
–4 –2 0 2 4 –2 y = 3 – 1 = 1
–1 = 2(8) + c 3 3
c = –17 Therefore the point of intersection is
–2

1
Therefore, y = 2x – 17 = 1 , 1 2 
3
3
1
–4 (c) y = x – 1
(c) 2y = 3x + 10 2 (b) 2y = 3x – 2 ................(1)
3
y = x + 5 x y y = 2x + 1 ................(2)
2
0 –1
3 (2) × 2
m =
2 2 0 2y = 4x + 2 ................(3)
new equation y = –x + 2 (1) – (3)
3 0 = –x – 4
y = x + c
2 x y x = –4
at the point(–2, –1) 0 2 Substitute x = –4 into (2)
3 2 0
–1 = (–2) + c y = 2(–4) + 1
2 y = –7
c = 2
Therefore the point of intersection is
3
Therefore, y = x + 2 = (–4, –7)
2

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BOOKLET ANS MATH F3.indd 26 03/01/2020 10:21 AM