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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
1 7 2 – (–2) 4 Substitute x = 0, y = 4
(c) y = x – 3 ..................(1) substitute x = into (3) (b) Gradient CD = =
2 2 4 – (–1) 5 2
4y = 6 – x ...................(2) 7 y – 0 4 4 = (0) + c
y = 8   – 26 Gradient AB = = 3
(1) × 4 2 3 – (–3) 5 c = 4
y = 2 y 4 2
4y = 2x – 12 ...............(3) = y = x + 4
6 5 3
2
4
(2) – (3) Therefore, the point of intersection is y = 4 4 m QR = – = –
0 = –3x + 18 =  7 , 2  4 5 6 3
x = 6 2 y = 5 x + c Equation of QR is :
2
Substitute x = 6 into (1) 1 16 At the point (–3, 0 ) y = – x + c
3
1 (c) y = x – ....................(1) substitute x = –3 and y = 0 2
y = (6) – 3 = 0 5 5 at the point (0, 4), c = 4, y = – x + 4
2 4 3
–3x + 7y = –8 ..................(2) 0 = (–3) + c
5
Therefore, the point of intersection is 12 (e) R(–3, 3)
Substitute (1) into (2) c =
= (6 , 0) 5 m RP = 0
–3x + 7  1 x – 16  = –8 Therefore equation of AB is Equation of RP is y = 3
14. (a) 7x – 4y = –7 ..........(1) 5 5
5x + y = 22 ............(2) 7 112 y = 4 x + 12 (f) Q(4, 0)
–3x + x – = –8 5 5
From (2) 5 5 and the equation of AB in general form m PQ = 8 = –2
y = 22 – 5x ............(3) 8 72 is 0 – 4
– x = 5 m RS = m PQ = –2
5
Substitute (3) into (1)  y = 4 x + 12  × 5
7x – 4(22 – 5x) = –7 x = –9 5 5 Equaton of RS:
7x – 88 + 20x = –7 substitute x = –9 into (2) 5y = 4x + 12 y = –2x + c
27x = 81 4x – 5y + 12 = 0 Substitute x = –3, y = 8
x = 3 –3(–9) + 7y = –8
y = –5 6 – 1 8 = –2(–3) + c
Substitute x = 3 into (3) (c) Gradient m PQ = = 5 c = 8 – 6 = 2
y = 22 – 5(3) Therefore the point of intersection is m OR = m PQ = 5 7 – 6
y = 7 = (–9, –5) y = –2x + 2
Equation of OR : y = 5x + c
Therefore, the point of intersection is
= (3, 7) 15. (a) y = –2x + 5, Substitute x = 0, y = 0
c = 0
Gradient PQ = –2
(b) 4x – 7y = 0 .......................(1) y = 5x
Equation PQ, y = –2x + c
8x – y – 26 = 0 .................(2)
Substitute x = 2 and y = –1,
From (2) –1 = –2(2) + c (d) m PR = undefined
y = 8x – 26 ........................(3) c = 3 Equation of PR is x = 6
substitute (3) into (1) Therefore the equation is y = –2x + 3 Q (0,4)
4x – 7(8x – 26) = 0 m PQ = 8 – 4 = 2
4x – 56x +182 = 0 6 3 2
52x = 182 Equation of PQ is y = x + c
3
7
x =
2

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BOOKLET ANS MATH F3.indd 27 03/01/2020 10:21 AM