06[Anal Add Math CD]

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Additional Mathematics SPM Chapter 6

Penerbitan Pelangi Sdn. Bhd.

6. Midpoint ofAB= (3, 4)

1 + 52

, t+ 22

= (3, 4) t+ 2

2= 4

t+ 2 = 8

t= 6

7. Midpoint of PQ= (1, 3)

2 + r2

, t 42

= (1, 3) 2 + r

2= 1 and t 4

2= 3

t 4 = 6

t= 10

2 + r= 2

r= 0

8. PQ= QR,

that is, Q(s, t) is the midpoint of PR.

1 + 32 , 4 62 = (s, t)

s= 1 + 32

and t = 4 62

= 1 = 1

9. 1 + x2 ,2 + y

2 = (4, 2)1 + x

2= 4 and 2 + y

2= 2

2 +y= 4

y= 6

1 +x= 8

x= 9

The coordinates of Care (9, 6).

10. (p, q) = 1 + 0.22 ,12

+ 4

2

= 0.4, 94Hence,p= 0.4, q=

94

11.2

1

A(2, 4)

P(x, y)

B(6, 10)

(x,y) = nx1+ mx2m+ n ,ny

1+ my

2m+ n

= 2(2) + 1(6)1 + 2 ,2(4) + 1(10)

1 + 2

= 103 , 6

The coordinates of Pare 103 , 6.

12. (a)

2

1

A(1, 0)

P(x, y)

B(4, 5)

(x,y) = 1(1) + 2(4)2 + 1 ,1(0) + 2(5)

2 + 1

= 3, 103 The coordinates of Pare 3, 103.(b)

2

3

A(1, 5)

P(x, y)

B(3, 1)

(x,y) = 3(1) + 2(3)2 + 3 ,3(5) + 2(1)

2 + 3 = 35 , 135 The coordinates of Pare 35 ,

13

5.(c)

2

1 1

2

B(6, 3)

P(x, y)

A(, 4)

(x,y) = 1(6) + 212

1 + 2

,1(3) + 2(4)

1 + 2 = 53 ,

11

3

The coordinates of Pare 53 ,11

3.

(d)21

21

1B(, 0)

P(x, y)

A(3, 2)

(x,y) = 1

12+

12 (3)

12

+ 1

,1(0) +

12 (2)

12

+ 1 = 23 ,

23

The coordinates of Pare 23 , 23.


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13.QR

RS

=13

3QR=RS

QR: QS= 1 : 2

2

1

S(2, 5)

Q(x, y)R(1, 4)

(x,y) = 2(1) + 1(2)1 + 2 ,2(4) + 1(5)

1 + 2

= 0, 133The coordinates of Qare 0, 133.

14. PS=13RS

PS

RS

=1

3PS: PR= 1 : 2

2

21

1

R(, 4)

P(x, y)

S(0, 8)

(x,y) = 2(0) + 112

1 + 2

,2(8) + 1(4)

1 + 2

= 16 , 4The coordinates of Pare 16 , 4.

15.2

1

B(x, y)

Q(2, 3)

A(1, 5)

(2, 3) = 1(1) + 2(x)2 + 1 ,1(5) + 2(y)

2 + 1

1 + 2x

3= 2 and

5 + 2y

3

= 3

y= 2

x=72

The coordinates ofBare 72 , 2.

16. (a) Area of ABC=12

0 1 3 0

4 2 5 4 =

12(0 + 5 + 12) (4 + 6 + 0)

=1217 10

= 72

unit2

(b) Area of ABC

=12

1 4 5 1

3 2 6 3 =

12(2 + 24 + 15) (12 10 6)

=1

241 + 4

=45

2

unit2

(c) Area of ABC

=12

0 4 2 0

1 3 5 1 =

12(0 20 + 2) (4 + 6 + 0)

=1218 2

=1

2

20

=12

(20)

= 10 unit2

(d) Area of ABC

=12

1 2 3 1

2 4 6 2 =

12(4 + 12 + 6) (4 + 12 + 6)

= 0 unit2

17. (a) Area ofABCD

=12

1 4 3 2 1

2 5 6 3 2 =

12(5 + 24 + 9 + 4) (8 + 15 + 12 + 3)

=1242 38

= 2 unit2

(b) Area ofABCD

=12

1 2 3 2 1

1 1 5 7 1 =

12(1 + 10 + 21 + 2) (2 3 10 7)

=1234 + 22

= 28 unit2


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(c) Area ofABCD

=12

0 1 2 3 0

3 4 1 1 3 =

12(0 + 1 + 2 9) (3 8 + 3 0)

=1

26 + 8

=122

= 1 unit2

(d) Area ofABCD

=12

0 1 2 3 0

1 3 5 7 1 =

12(0 + 5 + 14 + 3) (1 + 6 + 15 + 0)

=1222 22

= 0 unit2

18. Area of PQR=12

1 2 3 1

3 6 9 3 =

12(6 + 18 + 9) (6 + 18 + 9)

= 0 unit2

Since the area is zero, therefore the points P, Qand

Rare collinear.

19. Area of OBC=13

2

12

0 3 x 0

0 2 5 0 = 132(0 + 15 + 0) (0 + 2x+ 0)= 13

15 2x= 13

15 2x= 13 or 15 2x= 13

2x= 15 13 2x= 15 + 13

x= 1 x= 14

20. Area of PQRS

=

1

2

0 1 2 3 0

1 4 7 10 1

=12(0 + 7 + 20 + 3) (1 + 8 + 21 + 0)

=1230 30

= 0 unit2

Since the area is zero, therefore P, Q, R and S are

collinear.

21. (a) y= 2x+ 1

Whenx= 0, y= 2(0) + 1

= 1

Wheny= 0, 0 = 2x+ 1

x= 12

x-intercept = 12

;y-intercept = 1.

(b) 2x y+ 3 = 0

Whenx= 0, 0 y+ 3 = 0

y= 3

Wheny = 0, 2x 0 + 3 = 0

x= 32

x-intercept = 32

;y-intercept = 3.

(c)x2

+y3

= 2

x

4

+y

6

= 1

x-intercept = 4;y-intercept = 6

22. (a) Gradient =6 4

2 3

= 2

(b) Gradient =2 5

4 3

= 3

(c) Gradient = 4 2

3 (1)

= 64

= 32

(d) Gradient =3 0

4 (5)

= 3

23. (a) Gradient = y-intercept

x-intercept

= 32

(b) Gradient =

2

3

=23

(c) Gradient = 4

23

= 4 32

= 6


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(d)x2

y5

= 2

x4

y

10

= 1

Gradient = 10

4

=

5

2

(e)x

4

+y3

=12

2 x4 +y3= 2

12

x

2

+2y

3

= 1

Gradient =

32

2

=34

24. (a) The equation of the straight line is

y 2 = 4(x 1)

y= 4x 4 + 2

y= 4x 2

(b) The equation of the straight line is

y 3 = 4(x+ 1)

y= 4x 4 + 3

y= 4x 1

(c) The equation of the straight line is

y+ 6 =14

(x 2)

y= 14x 1

2 6

y=14x

13

2

25. (a) The equation of lineABis

y 1x 2

=4 1

3 2

= 3

y 1 = 3(x 2)

= 3x 6

3x y 5 = 0

(b) The equation of lineABis

y (3)

x (2)

=5 (3)

1 (2)

y+ 3x+ 2

= 2

y+ 3 = 2(x+ 2)

= 2x 4

2x+ y+ 7 = 0

(c) The equation of lineABis

y 5

x (1)

=2 5

0 (1)

y 5x+ 1

= 7

y 5 = 7(x+ 1)

= 7x 7 7x+ y+ 2 = 0

26. (a) The equation of the straight line is

x

x-intercept

+y

y-intercept

= 1

x3

+y4

= 1

(b)x

3

+y

1

= 1

x3

y= 1

(c) x1 +y

2 = 1

xy2

= 1

(d)x

12

+y

4

= 1

2xy4

= 1

27. (a) y= 3x+ 1

Gradient, m= 3

y-intercept = 1

Wheny= 0, 0 = 3x+ 1

x= 13

x-intercept = 13

(b) 2y= 4x 3

y= 2x32

Gradient, m= 2

y-intercept = 32

Wheny= 0, 2x= 32

x= 34

x-intercept = 34

(c) 2x+ y= 5

y= 2x+ 5

Gradient , m = 2

y-intercept = 5

Wheny= 0, 2x= 5

x=52

x-intercept =

5

2


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(d) 2y12x+ 5 = 0

2y=12x 5

y=14x

52

Gradient,m=14

y-intercept = 52

Wheny= 0,14x=

52

x= 10

x-intercept = 10

(e)x2

+y3

= 1

Gradient , m = 32

x-intercept = 2 y-intercept = 3

(f)12x

13y+ 4 = 0

12x

13y= 4

12x

4

13y

4

= 4 4

x8

+y

12

= 1

Gradient,m =

12

8

=32

x-intercept = 8 y-intercept = 12

28. (a) 2y= 3x 1

3x 2y 1 = 0

(b)x2

=y3

+ 1

6x2= 6y3

+ 1 3x= 2y+ 6

3x 2y 6 = 0

(c)x+ 1

3=

y4

4(x+ 1) = 3y

4x+ 4 = 3y

4x 3y+ 4 = 0

29. (a) y= 3x 1 ....................... y= 4x+ 5 .......................

= , 3x 1 = 4x+ 5 4x 3x= 1 5

x= 6

Substitutex= 6 into ,

y= 3(6) 1 = 19

Point of intersection = (6, 19)

(b) x+ 2y= 1 ...................................

x2

4 = 3y..................................

2, x 8 = 6y x 6y= 8 ....................

, 8y= 7

y= 78

Substitutey= 78

into ,

x+ 2 78= 1 x= 1 +

74

=11

4

Point of intersection = 114 , 78

(c) 2x+ 3y= 5 .................................. 6x 2y= 1 ................................

3, 6x+ 9y= 15 ................ , 11y= 16

y=16

11

Substitutey=16

11

into ,

2x+ 3 1611= 5 2x= 5

48

11

=7

11

x=7

22 Point of intersection 722 ,

16

11

30. (a) y= 2x 1

Gradient = 2

2y= 4x+ 3

y= 2x+32

Gradient = 2

Hence, the two lines are parallel.


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(b) 3x y+ 4 = 0

y= 3x+ 4

Gradient = 3

3x+ y 5 = 0

y= 3x+ 5

Gradient = 3

Hence, the two lines are not parallel.

(c)x2

+y3

= 1

Gradient = 32

2y= 3x 5

y= 32x

52

Gradient = 32

Hence, the two lines are parallel.

31. (a) y= 3x 1

Gradient = 3

y= kx+ 4

Gradient = k

Since the two lines are parallel,

k= 3

(b) y= 4x+ 3

Gradient = 4

y=k

2x 5

Gradient =k

2 Since the two lines are parallel,

k

2

= 4

k= 8

(c) x+ 2y= 4

y= 12x+ 2

Gradient = 12

y 2kx+ 3 = 0

y= 2kx 3

Gradient = 2k


2k= 12

k= 14

(d)x2

+y4

= 0

Gradient = 42

= 2

3y kx 4 = 0

3y= kx+ 4

y=k

3x+

43

Gradient =k

3


k3

= 2

k= 6

32. (a) y= 3x 6

Gradient = 3 The equation for the parallel line is

y 2 = 3(x 1)

y= 3x 3 + 2

y= 3x 1

(b) 2y= 4x+ 3

y= 2x+32

Gradient = 2

The equation for the parallel line is

y 3 = 2(x+ 1)

y= 2x+ 2 + 3

y= 2x+ 5

(c) 4x y+ 1 = 0

y= 4x+ 1

Gradient = 4


y+ 2 = 4(x 0)

y= 4x 2

(d)x2

y6

= 1

Gradient = 6

2

= 3


y + 3 = 3(x+ 1)

y= 3x+ 3 3

y= 3x

33. (a) y= 4x 1 Gradient = 4

y= 14x+ 3

Gradient = 14

m1m

2= (4) 14

= 1

The two lines are perpendicular.


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(b) 2y= 6x+ 5

y= 3x+52

Gradient = 3

y=13x 4

Gradient =

1

3

m1m

2= (3) 13

= 1


(c) x+ 2y= 5

2y= x+ 5

y= 12x+

52

Gradient = 12

2y 4x= 7

2y= 4x+ 7

y= 2x+72

Gradient = 2

m1m

2= 12(2)

= 1


(d) x y= 8

y=x 8

Gradient = 1

2x+ y= 1 y= 2x+ 1

Gradient = 2

m1m

2= (1)(2)

= 2

The two lines are not perpendicular.

(e)x2

y4

= 1

Gradient = 42

= 2

3y= x+ 6

y= 13x+ 2

Gradient = 13

m1m

2= (2) 13

= 23

The two lines are not perpendicular.

34. (a) y= kx 1

Gradient = k

y= 4x+ 3

Gradient = 4

m1m

2= 1

(4)(k) = 1

k= 14

(b) 2x+ ky= 1

ky= 2x+ 1

y= 2kx+

1k

Gradient = 2k

y=16x 1

Gradient =16

m1m

2= 1

2k16= 1

1

3k

= 1

3k= 1

k=13

(c) 2y+ 4kx= 3

2y= 4kx+ 3

y= 2kx+32

Gradient = 2k

x2

+y6

= 1

Gradient = 62

= 3

m1m

2= 1

(2k)(3) = 1

6k= 1

k= 16

(d)12kx+ 2y= 5

2y= 12kx+ 5

y= 14kx+

52

Gradient = 14k

4x+ 3y= 6

3y= 4x+ 6

y= 43x+ 2

Gradient = 43


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m1m

2= 1

14k43= 1

k

3

= 1

k= 3

35. (a) y= 4x 1

Gradient = 4

The equation of the perpendicular line is

y 3 = 14

(x 1)

y= 14x+

14

+ 3

y= 14x+

134

(b) y= 12x+ 4

Gradient = 12 The equation of the perpendicular line is

y 2 = 2(x+ 1)

y= 2x+ 2 + 2

y= 2x+ 4

(c) 2xy= 2

y= 2x 2

Gradient = 2


y+ 3 = 12

(x 0)

y= 12x 3

(d)x3

+y4

= 1

Gradient = 43


y + 2 =34

(x+ 1)

y=34x+

34

2

y=

3

4x

5

4

36. y= 2x 1............................................y= 4x+ 3 ...........................................

= , 2x 1 = 4x+ 3 2x= 4

x= 2

Substitutex= 2 into ,y= 2(2) 1

= 5

Point of intersection = (2, 5)

The equation of the line is

y+ 5 = 3(x+ 2)

y= 3x+ 6 5

y= 3x+ 1

37. 2x y= 4

y= 2x 4

Gradient = 2


y 2 = 2(x+ 1)

y= 2x+ 2 + 2

y= 2x+ 4

38. Gradient ofAB=6 (3)

5 (1)

=96

=3

2Gradient of PQ=

23

The equation of line PQis

y 6 = 23

(x 5)

y= 23x+

103

+ 6

y= 23x+

283

39. (a) The equation of locus is

(x 0)2+ (y 0)2= 2 x2+ y2= 4 x2+ y2 4 = 0

(b) The equation of locus is

(x 1)2+ (y 2)2 = 3 (x 1)2+ (y 2)2= 9

x2 2x+ 1 +y2 4y+ 4 9 = 0

x2+ y2 2x 4y 4 = 0

(c) The equation of locus is

(x+ 1)2+ (y 3)2= 4 (x+ 1)2+ (y 3)2= 16

x2+ 2x+ 1 +y2 6y+ 9 16 = 0 x2+ y2+ 2x 6y 6 = 0

40. (a)PAPB

= 1

PA= PB

(x 0)2+ (y 1)2 = (x 2)2+ (y 3)2 x2+ (y 1)2= (x 2)2+ (y 3)2

x2+ y2 2y+ 1 =x2 4x+ 4 +y2 6y+ 9

4x+ 4y 12 = 0

x+ y 3 = 0

Hence, the equation of locus isx+ y 3 = 0.


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7. (a) The equation of PQis x4

+y8

= 1.

(b)

3

1

P(4, 0)

Q(0, 8)S(x, y)

(x,y) = 1(4) + 3(0)3 + 1

,1(0) + 3(8)

3 + 1)

= (1, 6)

The coordinates of Sare (1, 6).

(c) x4

+y8

= 1

Gradient of PQ = 8

4

= 2

Gradient ofRS = 12

Let the coordinates ofRbe (x1, 0).

0 6

x

1 (1)

= 12

6

x

1 (1)

= 12

x1+ 1 = 12

x1= 11

Hence, thex-intercept ofRSis 11.

8. (a) (i) Radius of the circle = (6 3)2+ (0 2)2

= 9 + 4

= 13 units PB= 13 (x 3)2+ (y 2)2 = 13 (x 3)2+ (y 2)2= 13

x2 6x+ 9 +y2 4y+ 4 13 = 0

x2+ y2 6x 4y= 0

The equation of the locus of point Pis

x2+ y2 6x 4y= 0.

(ii) SubstituteD(t, 4) into the equation of locus,

t2+ 42 6t 4(4) = 0

t2 6t= 0

t(t 6) = 0 t = 0 or t 6 = 0

t= 6

(b)

O

E(0, y1)

C(6, 0)

B(3, 2)

y

x

Gradient ofBC =2 0

3 6

= 23

Gradient of CE=32

Let the coordinates ofEbe (0,y1).

y

1 0

0 6

=32

y1=

32

(6)

y1= 9

Area of COE=12

6 9

= 27 unit2

9. (a) (i) x+ 2y 6 = 0

2y= x+ 6

y= 12x+ 3

Gradient of PQ= 12

Gradient ofRQ= 2

The equation of lineRQis

y+ 3 = 2(x 1)

y= 2x 2 3

y= 2x 5

(ii) y= 2x 5 .................. x + 2y 6 = 0 ...........................

Substitute into , x+ 2(2x 5) 6 = 0 x+ 4x 10 6 = 0

5x= 16

x=16

5

Substitutex=16

5

into ,

y= 2 165 5

=75

The coordinates of Qare 165

,75.

(b)

3

55716

2

R(1, 3)

Q(,)

S(x, y)

165

,75 =

3(1) + 2x

2 + 3,

3(3) + 2y

2 + 3

= 3 + 2x5

,2y 9

5


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16

5

=3 + 2x

5

and75

=2y 9

5

2y 9 = 7

y= 8

3 + 2x = 16

x =13

2

The coordinates of Sare 132

, 8.

(c) RM= 3

(x 1)2+ (y+ 3)2 = 3 (x 1)2+ (y+ 3)2= 9

x2 2x+ 1 +y2+ 6y+ 9 = 9

x2+ y2 2x+ 6y+ 1 = 0

The equation of the locus of pointMis

x2+ y2 2x+ 6y+ 1 = 0.

10. (a) Area of ABC

=1

2

0 2 2 0

3 1 4 3

=12(0 + 8 + 6) (6 + 2 + 0)

=1214 + 4

= 9 unit2

(b) D = 3(2) + 1(2)1 + 3 ,3(4) + 1(1)

1 + 3

= 1, 114

(c) (i) PA = 2PC

(x+ 2)2+ (y 4)2= 2(x 2)2+ (y+ 1)2 (x+ 2)2+ (y 4)2= 4[(x 2)2+ (y+ 1)2]

x2+ 4x+ 4 +y2 8y+ 16

= 4[x2 4x+ 4 +y2+ 2y+ 1]

= 4x2 16x+ 16 + 4y2+ 8y+ 4

3x2+ 3y2 20x+ 16y= 0


3x2+ 3y2 20x+ 16y= 0.

(ii) Assume the locus intersects the x-axis,

substitutey= 0 into the equation of locus.

3x2 20x= 0

x(3x 20) = 0

x= 0,x=20

3

Hence, the locus intersects thex-axis at two

points.

1. AB= (5 1)2+ (5 2)2

= 16 + 9 = 5 units

AB= 2BC

BC=52

units

2. AB= 16

(k+ 1)2+ (4 3)2= 16 (k+ 1)2+ 1 = 256 (k+ 1)2= 255

k+ 1 = 255

k= 255 1

= 255 1, 255 1

3. Eis the midpoint ofAC.

E= 1 + 72 ,2 + 6

2 = (4, 4)

4.

1

2

A(2, 0)

B(0, 4)

C(x, y)

AB: AC= 1 : 3

AB: BC= 1 : 2

(0, 4) = 2(2) + 1(x)1 + 2

,2(0) + 1(y)

1 + 2

= x 43

,y3

x 4

3= 0 and

y3

= 4

x= 4 y= 12


5. Let the coordinates ofDbe (0,y).

Gradient of CD= Gradient ofAC

y 60 3

=6 1

3 (2)

y 6 = 3(1) y= 3

Area of BCD =12

0 5 3 0

3 2 6 3 =

12(0 + 30 + 9) (15 + 6 + 0)

=1239 21

= 9 unit2


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6. Area of quadrilateral PQRS

=12

0 5 2 1 0

3 2 6 1 3=

12(0 + 30 + 2 + 3) (15 4 6 + 0)

=1

2

35 + 25

= 30 unit2

7. Area of ABC= 16

12

1 0 k 1

2 3 4 2 = 16(3 + 0 + 2k) (0 + 3k 4)= 32

1 k= 32 1 k= 32 or 1 k= 32

k= 31 k= 33

8. (a) Gradient = 5 (1)3 (3)

= 1

The equation of lineABCDis

y 5 = 1(x 3)

y=x 3 + 5

y=x+ 2

(b) y-intercept = 2

Wheny= 0, 0 =x+ 2

x= 2

x-intercept = 2

9. (a) Gradient ofRQ= 2 Gradient of PQ=

12

The equation of PQis

y+ 1 =12

(x+ 4)

y=12x+ 2 1

y=12x+ 1

(b) Fory= 2x+ 1,

wheny= 0, 0 = 2x+ 1

x= 12

Thex-intercept ofRQis12

.

10. (a) 2x y= 4

y= 2x 4

Gradient of CD= 2

Gradient ofAB= 2

The equation of lineABis

y 5 = 2(x 2)

y= 2x 4 + 5

y= 2x+ 1

(b) y= x 2 ............................ 2x y= 4 ....................................

Substitute into , 2x (x 2) = 4

2x+ x+ 2 = 4

3x= 2

x=23

Substitutex=23

into ,

y= 23 2

= 83

The coordinates ofDare 23 , 83.

11. (a) PA= 5

(x+ 1)2

+ (y 2)2

= 5 (x+ 1)2+ (y 2)2= 25

x2+ 2x+ 1 +y2 4y+ 4 25 = 0

x2+ y2+ 2x 4y 20 = 0


x2+ y2+ 2x 4y 20 = 0.

(b) Substitutex= 2 andy= kinto the equation,

4 + k2+ 2(2) 4k 20 = 0

k2 4k 12 = 0

(k 6)(k+ 2) = 0

k 6 = 0 or k+ 2 = 0

k= 6 k= 2

12. AP: PB= 2 : 3

APPB

=23

3AP= 2PB

3(x 1)2+ (y 4)2= 2(x 3)2+ (y+ 2)29[(x 1)2+ (y 4)2] = 4[(x 3)2+ (y+ 2)2]

9(x2 2x+ 1 +y2 8y+ 16)

= 4(x2 6x+ 9 +y2+ 4y+ 4)

9x2 18x+ 9 + 9y2 72y+ 144

= 4x2

24x+ 36 + 4y2

+ 16y+ 165x2+ 5y2+ 6x 88y+ 101 = 0


5x2+ 5y2+ 6x 88y+ 101 = 0.

13. (a) Substitutex= 1 andy= kintox2+ y2= 4,

1 + k2= 4

k2= 3

k= 3


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15



(b) Gradient of OA =3 0

1 0

= 3

Gradient of tangent atA= 1

3

The equation of the tangent atAis

y 3 = 13

(x 1)

y= 1

3

x+1

3

+ 3

y= 1

3

x+4

3

14. (a) Let the coordinates of Cbe (x,y).

(4, 0) = 2 + x2 ,2 + y

2

2 + x

2

= 4 and2 + y

2= 0

x= 6 y= 2


(b) Gradient ofBC=0 (2)

4 2

= 1

Gradient ofAD= 1

The equation of lineADis

y 0 = 1(x 4)

y= x+ 4

(c) Let the point of intersection of BCat the y-axis

beE(0,y).

Gradient ofBD= Gradient ofBE

1 =y (2)

0 2 2 =y+ 2

y= 4

They-intercept of lineBCis 4.

15. (a) Area of PQR

=12

3 2 6 3

5 3 1 5 =

12(9 2 + 30) (10 + 18 + 3)

=1237 11

= 13 unit2

(b) Let the intersection of line PQand they-axis be

S(0,y1).

Gradient of PS= Gradient of PQ

y

1 5

0 3

=5 3

3 (2)

y1 5 =

25

(3)

y1=

65

+ 5

=19

5

They-intercept of line PQis19

5

.

(c) QM=12MR

QM: MR= 1 : 2

1

2

Q(2, 3)

R(6, 1)

M(x, y)

(x,y) = 2(2) + 1(6)1 + 2 ,2(3) + 1(1)

1 + 2

= 23 ,73

The coordinates ofMare 23 ,73.

16. (a) Let the intersection of lineBCand they-axis beE(0,y).

Gradient ofBE= Gradient ofBC

y 40 3

=4 (8)

3 (1)

y 4 = 3 124

= 9

y= 5

They-intercept of lineBCis 5.

(b) Gradient ofAD= Gradient ofBC

=4 (8)

3 (1)

= 3

The equation of lineADis

y 6 = 3(x+ 3)

= 3x+ 9

y= 3x+ 15


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16



(c) Let the coordinates ofDbe (x,y).

Midpoint ofBD= Midpoint ofAC

3 + x2

,4 + y

2

= 3 + (1)2

,6 + (8)

2

= (2, 1)

3 + x

2

= 2 and4 + y

2

= 1

x= 7 4 +y= 2

y= 6

The coordinates ofDare (7, 6).

(d) Area of rectangleABCD

=12

3 3 7 1 3

4 6 6 8 4 =

12(18 + 18 + 56 4) (12 42 + 6 24)

=1288 + 72

= 80 unit2

17. (a) Gradient ofBC= Gradient of CD

=0 (3)

2 (1)

= 1

The equation of lineBCis

y 0 = 1(x 2)

y=x 2 ......... Equation ofAB,x 2y+ 4 = 0 ...............

Substitute into , x 2(x 2) + 4 = 0

x+ 8 = 0

x= 8

Substitutex= 8 into , y= 8 2

= 6

The coordinates ofBare (8, 6).

(b)

3

2C(1, 3)

B(8, 6)

E(x, y)

(1, 3) = 3x+ 2(8)2 + 3

,3y+ 2(6)

2 + 3

= 3x+ 165

,3y+ 12

5

3x+ 16

5

= 1 and3y+ 12

5

= 3

3x= 21 3y= 27

x= 7 y= 9

The coordinates ofEare (7, 9).

(c) x 2y+ 4 = 0

Whenx= 0, 2y+ 4 = 0

y= 2

F(0, 2)

Area of BCF

=

12

0 1 8 0

2 3 6 2

=12(0 6 + 16) (2 24 + 0)

=1210 + 26

= 18 unit2

18. (a) y= 2x+ 6

Gradient ofAB= 2

Gradient of CD= 2

The equation of line CDis

y+ 3 = 2(x 1) = 2x+ 2

y= 2x 1

(b) Substitutex = 2 andy= kintoy= 2x+ 6,

k= 2(2) + 6

k= 2

Gradient of CE= Gradient ofBC

0 (3)

p 1

=2 (3)

2 1

3 = 5(p 1)

p 1 =35

p=85

(c)

n

m

E(, 0)58

C(1, 3)

B(2, 2)

Usey-coordinate,

(3)n+ 2m

m+ n= 0

2m 3n= 0

2m= 3n

mn =

32

CE: EB= 3 : 2

(d) Area of BOC =12

0 1 2 0

0 3 2 0 =

12(0 + 2 + 0) (0 6 + 0)

=122 + 6

= 4 unit2


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17



19. (a) PA: PB= 1 : 2

PA

PB

=12

PB= 2PA

(x 2)2+ (y 0)2= 2(x 0)2+ (y 1)2 (x 2)2+ y2= 4[x2+ (y 1)2]

x2

4x+ 4 +y2

= 4(x2

+ y2

2y+ 1) = 4x2+ 4y2 8y+ 4

3x2+ 3y2+ 4x 8y= 0

(b) Substitutex= 43

andy = 0 into

3x2+ 3y2+ 4x 8y= 0,

LHS = 3x2+ 3y2+ 4x 8y

= 3 432

+ 3(0)2+ 4 43 8(0) =

163

163

= 0

= RHS

Hence, the point 43 , 0lies on the locus of P.

(c) Substitutey= 0 into 3x2+ 3y2+ 4x 8y= 0,

3x2+ 4x= 0

x(3x+ 4) = 0

x= 0 or 3x+ 4 = 0

x= 43

The points of intersection are (0, 0) and (43

, 0).

(d) Substitutex= 0 into 3x2

+ 3y2

+ 4x 8y= 0, 3y2 8y= 0

y(3y 8) = 0

y= 0 or 3y 8 = 0

y=83

Since there are values fory-coordinate, then the

locus intersects they-axis.

20. (a) Area of ABC

=12

1 8 4 1

2 3 7 2

=12(3 + 56 + 8) (16 + 12 7)

=1261 21

= 20 unit2

Let dbe the perpendicular distance fromBto line

AC.

Distance ofAC = [(8 (1)]2+ (3 2)2

= 81 + 1 = 82 units

Area of ABC= 20

12

d82 = 20

d=40

82

= 4.417 units

(b)

P(1, 3)

Q(h, k)

Midpoint of PQ= 1 + h2

,3 + k

2

Since the midpoint of PQlies on the perpendicular

bisector, so we substitute x =1 + h

2 and

y=3 + k

2

into 3x+ 5y 16 = 0,

31 + h2

+ 53 + k2

16 = 0

3 + 3h

2+

15 + 5k

2 16 = 0

3 + 3h 15 + 5k 32 = 0

3h+ 5k= 50 ......... 3x+ 5y 16 = 0

5y= 3x+ 16

y= 35x+

16

5

Gradient of perpendicular bisector = 3

5 Gradient of line PQ=

53


y+ 3 =53

(x+ 1)

=53x+

53

y=53x

43

Substitutex= h,y= kinto the equation of PQ,

k=53h

43

................................

Substitute into ,

3h+ 5 53h43= 50

3h+25

3

h20

3

= 50

33h+ 253

h20

3= 3(50)

9h+ 25h 20 = 150

34h= 170

h= 5


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18



Substitute h= 5 into , 3(5) + 5k= 50

k= 7

21. (a)PA

PB

=12

PB= 2PA (x 0)2+ (y+ 2)2 = 2(x 0)2+ (y 1)2 x2+ (y+ 2)2= 4[x2+ (y 1)2]

x2+ y2+ 4y+ 4 = 4(x2+ y2 2y+ 1)

= 4x2+ 4y2 8y+ 4

3x2+ 3y2 12y= 0

x2+ y2 4y= 0


x2+ y2 4y= 0.

(b) Substitutex= 2 andy= 2 intox2+ y2 4y= 0,

LHS =x2+ y2 4y

= 22+ 22 4(2)

= 0

= RHS

Hence, C(2, 2) lies on the locus of point P.

(c) Gradient ofAC=2 1

2 0

=12

Equation ofAC,y=12x+ 1 ...................

Equation of locus,x2+ y2 4y= 0 ..........

Substitute into ,

x2+ 12x+ 12

4 12x+ 1= 0 x2+

14x2+ x+ 1 2x 4 = 0

54x2 x 3 = 0

5x2 4x 12 = 0

(5x+ 6)(x 2) = 0

5x+ 6 = 0 or x 2 = 0

x= 65

x= 2

Substitutex= 65 into ,

y=12

65+ 1

= 35

+ 1

=25

The coordinates ofDare 65 ,25.

(d) Gradient ofAC=12

Gradient ofBD=

25

(2)

65

0

= 125 56 = 2

Gradient ofACGradient ofBD=12

(2)

= 1

Hence, linesACandBDare perpendicular to each

other.

22. (a) PQ = 10

(q 0)2+ (0 p)2= 10 p2+ q2= 100

(b) (i) RQ= 3PR

PR: RQ= 1 : 3

P(p, 0)

Q(0, q)3

1R(x, y)

(x,y) = 3p+ 01 + 3 , 0 + q1 + 3 =

3p

4,q4

3p

4

=x andq4=y

p=4x

3

q= 4y

Substitutep=4x

3

and q= 4yinto

q2+ p2= 100,

(4y)2+ 4x32

= 100

16y2+16

9 x2= 100

16

9 x2+ 16y2 100 = 0

The equation of the locus of pointRis

16

9 x2+ 16y2 100 = 0.


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19



(ii) Substitutey= 0 into16

9 x2+ 16y2 100 = 0,

16

9 x2 100 = 0

x2= 100 916

x=

90016

= 304

= 15

2

Thex-coordinate ofRis 15

2

.

23. (a) Gradient of PQGradient ofRQ= 1

5 21 4

t 2r 4

= 1

(1)t 2

r 4= 1 t 2 = r 4 t= r 2

(b) Area of PQR

=12

1 r 4 1

5 t 2 5 =

12(t+ 2r+ 20) (5r+ 4t+ 2)

=12

(t+ 2r+ 20 5r 4t 2)

=1

2

(3t 3r+ 18)

= 32t

32r+ 9

= 9 32

(r+ t)

(c) Given the area of rectangle PQRS= 30 unit2

Area of PQR= 15 unit2

9 32

(r+ t) = 15

32

(r+ t) = 6

r+ t= 4 ................

From (a), t= r 2 ............ Substituteinto , r+ r 2 = 4

2r= 2

r= 1

Substitute r= 1 into , t = 1 2

= 3

The coordinates ofRare (1, 3)

1. Substitutex= 2,y= tinto equationx2+ y2= 16,

22+ t2= 16

t2= 12

t= 12

Based on the diagram, t = 12

Gradient of OA=12 0

2 0

=12

2

=23

2

12 = 4 3 = 4 3 = 23

= 3Gradient of tangentABis

1

3Equation of tangentABis

y 12 = 13

(x 2)

y= 1

3

x+2

3

+ 12

= 1

3

x+2

3

+ 23

2. Let P(x,y)

Gradient of PQ= Gradient ofRS

y (1)

x (1)

=4 2

0 (2)

= 1

y+ 1 =x+ 1

y =x............................

mPS

mPQ

= 1

y 4x 0

y+ 1x+ 1

= 1

y 4x

y+ 1x+ 1

= 1 (y 4)(y + 1) = x(x+ 1)

y2 3y 4 = x2 x

y2 3y+ x2+ x 4 = 0......................

Substitute into ,x2 3x+ x2+ x 4 = 0

2x2 2x 4 = 0

x2 x 2 = 0

(x+ 1)(x 2) = 0

x= 1 orx= 2

Based on the diagram,x= 2


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20



Substitutex= 2 into ,y= 2

The coordinates of Pare (2, 2).

Area of trapezium PQRS

=1

2

0 2 1 2 0

4 2 1 2 4

=12

[(0 + 2 2 + 8) (8 2 2 + 0)]

=12

[8 (12)]

=12

(20)

= 10 unit2

3. Gradient ofAC = 3

k (2)

h (1)

= 3

k+ 2

h+ 1

= 3

k+ 2 = 3h+ 3

k = 3h+ 1 ....................

Gradient ofABGradient ofBC = 1

6 (2)

3 (1)

k 6h 3 = 1

2k 6h 3 = 1 2(k 6) = 1(h 3)

2k 12 = h+ 3 2k = h+ 15 .........

Substitute into ,2(3h+ 1) = h+ 15

6h+ 2 = h+ 15

7h = 13

h =13

7

Substitute h=13

7

into ,

k= 3 137+ 1

=39

7

+ 1

=46

7

4. (a) Area of ABC= 4

Since there are two possible positions for point

C,

therefore12

1 2 k 1

1 5 2k 1 = 4 [(5 + 4k k) (2 + 5k+ 2k)] = 8

5 + 3k (2 + 7k) = 8

5 + 3k+ 2 7k = 8

7 4k = 8

4k = 7 8

= 7 8 or 7 + 8

= 1 or 15

k = 14

or154

(b) Gradient ofABGradient ofBC = 1

5 (1)

2 1

2k+ 1k 1 = 1

6 2k+ 1k 1 = 1 6(2k+ 1) = 1(k 1)

12k+ 6 = k+ 1

13k = 5

k = 5

13

(c) Gradient ofAB = Gradient ofBC

5 (1)

2 1

=2k (1)

k 1

6 =2k+ 1

k 1

6k 6 = 2k+ 1

4k = 7

k =74

5.

(a) Midpoint of PQ = 4 + r

2 ,

9 + t

2 (b)

02y+ x= 7

7

2

7

y

A

P(4, 9)

Bx

Gradient of PQGradient ofAB= 1

t 9r 4

72

7= 1

t 9r 4

12= 1

t 9

r 4

= 2

t 9 = 2(r 4)

= 2r 8

t= 2r+ 1


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21



(c) If r= 2,

t= 2(2) + 1

= 3

PQ= [4 (2)]2+ [9 (3)]2

= 36 + 144 = 180 = 36 5 = 65 units

6. (a) (i)

O

D

C

EF

B

A(14, 0)

y

y + 3x 6 = 0

x

Whenx= 0,

y+ 3x 6 = 0

y+ 3(0) 6 = 0

y= 6

The coordinates ofEare (0, 6).

Wheny= 0,

y+ 3x 6 = 0

0 + 3x 6 = 0 x= 2

The coordinates of Fare (2, 0).

LetB= (x,y)

SinceEis the midpoint ofBF,

thenx+ 2

2= 0

x= 2

y+ 0

2= 6

y= 12 Therefore, the coordinates of B are

(2, 12).

(ii) Area of quadrilateral OABE

=120 0 2 14 00 6 12 0 0

=12

[(0 + 0 + 0 + 0) (0 12 168 + 0]

=12

180

= 90 unit2

(b)D(x, y)

B(2, 12)1

3

A(14, 0)

(1)x+ 3(14)

1 + 3

= 2

x 42

4= 2

x 42 = 8

x= 34

(1)y+ 3(0)

1 + 3

= 12

y = 4 12 = 48

The coordinates ofDare (34, 48).

(c) (i) mAC

= mAB

y 0

0 + 14

=12 0

2 + 14

y14

=12

12

y= 14


Let the moving point be P(x,y).

PE= 2PC

(x 0)2+ (y 6)2 = 2(x 0)2+ (y 14)2

x2+ (y 6)2= 4[x2+ (y 14)2] x2+ y2 12y+ 36 = 4(x2+ y2 28y+ 196)

= 4x2+ 4y2 112y+ 784

3x2+ 3y2 100y+ 748 = 0

(ii) At they-axis,x= 0

3y2 100y+ 748 = 0

b2 4ac= (100)2 4(3)(748)

= 1024 0

The locus intersects they-axis.

7. (a) y = 2x..........................................

y = 8x

..........................................

= ,

2x =8x

x2 = 4

x = 2

Based on the diagram,x= 2.

Substitutex= 2 into , y= 2(2)

= 4

The coordinates ofAare (2, 4).


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22



(b) LetB(x,y)

1(x) + 3(0)

1 + 3

= 2

x= 8

1(y) + 3(0)

1 + 3

= 4

y= 16

The coordinates ofBare (8, 16).

(c) The gradient of the perpendicular line is 12

.

The equation of the straight line is

y 16 = 12

(x 8)

= 12x+ 4

y = 12x+ 20

8. (a) Substitutey = 0 into equationy= 3x2 12, 3x2 12 = 0

3(x2 4) = 0

x2 4 = 0

x2 = 4

x = 2

Based on the graph, the coordinates of P are

(2, 0).

The coordinates of Qare (0, 19).

Gradient of PQ=0 (19)

2 0

= 192


y 0 =192

(x 2)

y =192 x 19

(b) Gradient of line PS= 2

19

The equation of line PSis

y 0 = 2

19

(x 2)

y= 2

19

x+4

19

(c) y= 2

19

x+4

19

.........................

y= 3x2 12 .................................

= ,

3x2 12 = 2

19

x+4

19

3x2+2

19

x23219

= 0

19, 57x2+ 2x 232 = 0

x=2 (2)2 4(57)(232)

2(57)

=2 52 900

114

= 116

57 , 2

x= 2 is ignored

because it is

x-coordinate for

point P.

Substitutex= 116

57

into ,

y = 2

19

11657+

419

=460

1083

The coordinates of Sare 11657 ,460

1083.

9. (a) (i) Gradient of PR=8 6

6 8

= 1

Gradient ofAC= 1 Since PR//AC

The equation of lineACis

y 12 = 1(x 10)

= x+ 10

y = x+ 22

(ii) The perpendicular bisector ofBCis PR.

Gradient of PR= 1

The equation of line PRis

y 6 = 1(x 8) y 6 = x+ 8

y= x+ 14

(b) Area of PQR

=12

6 10 8 6

8 12 6 8 =

12|[(72 + 60 + 64) (80 + 96 + 36)]|

= 8 unit2

Area of ABC= 22(8)

= 32 unit2

Area of PQR: Area of ABC

= 8 : 32

= 1 : 4

(c) LetB(x,y)

Since Pis the midpoint of AB

x+ 82

= 6 andy+ 14

2= 8

y= 2x= 4

Therefore, the coordinates ofBare (4, 2).


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23



10. (a) OA= 80 (2k)2+ k2 = 80 4k2+ k2= 80

5k2= 80

k2= 16

k= 4

Since k

0, therefore k= 4.

(b) x-coordinate ofB= 2k

= 2(4)

= 8

y-coordinate ofB= 42

GivenAC: CB

= 2 : 1

= 2


(c) Gradient of OB=2 0

8 0

= 14

The equation of OBis y= 14x.

11. (a) Wheny= 0,

y2= 6x+ 9

02= 6x+ 9

6x= 9

x= 96

= 3

2 The coordinates of Qare (

32

, 0).

Whenx= 0,

y2= 6(0) + 9

y2= 9

y= 3

The coordinates of Pare (0, 3).

The equation of PQis

y 3 =3 0

0 32(x 0)

y 3 = 2x

y= 2x+ 3

(b) Gradient of QS= 12

mPQ

mQS

= 1

The equation of line QSis

y 0 = 12x+

32

y= 12x

34

(c) y= 12x

34

............................

y2= 6x+ 9 ...................................

Substitute into ,

12x34

2

= 6x+ 9

14x2+ 2 12x

34+

34

2

= 6x+ 9

14x2+

34x+

916

= 6x+ 9

16 14x2+

34x+

916= 16(6x+ 9)

4x2+ 12x+ 9 = 96x+ 144

4x2+ 12x+ 9 96x 144 = 0

4x2 84x 135 = 0

(2x+ 3)(2x 45) = 0

2x 45 = 0

x =45

2 Substitutex= 452

into ,

y = 12

452

34

= 454

34

= 12

Therefore, the coordinates of Sare 452 , 12.

12. (a) Since PQRSis a parallelogram,

Midpoint of PR= Midpoint of QS

h+ 62 , 2k 52 = 2h 12 , k+ 1 + 42

h+ 6

2 =

2h 1

2 and

2k 5

2=

k+ 5

2 h+ 6 = 2h 1 2k 5 = k+ 5

h = 7 k= 10

(b) P(7, 20), Q(14, 11),R(6, 5), S(1, 4)

Let T(x,y) be the point of intersection of diagonals

PRand QS.

T(x,y) = Midpoint of PR

x=6 + h

2

=6 + 7

2

=132

y=2k 5

2

=20 5

2

=152

Therefore, the point of intersection of diagonals

PRand QSis T132 ,

152.

x= 3

2

is ignored

because it is

x-coordinate of Q.


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24



(c) Gradient of QR=11 + 5

14 6

=168

= 2

The equation of line passing through T and is

parallel to QRis

y152

= 2x 132 = 2x 13

y = 2x 13 +152

y = 2x11

2

13. (a) A= 1 + 32

,8 10

2

= (2, 1)

(b) Midpoint of PR= Midpoint of QS

1 + 32

,8 10

2

= h 42

,k+ 5

2

(2, 1) = h 4

2,k+ 5

2

h 4

2

= 2 andk+ 5

2

= 1

h 4 = 4 k+ 5 = 2

h = 8 k = 7

(c) P(1, 8), S(4, 5)

Gradient of PS=8 5

1 ( 4) =

35

The equation of the line passing through Aand

parallel to PSis

y (1) =35

(x 2)

y+ 1 =35x

65

y=35x

65

1

y=

3

5x

11

5

14. (a) ForB, substitutey= 0 into 3y 4x+ 12 = 0,

4x+ 12 = 0

x= 3


ForA, substitutex= 0 into 3y 4x+ 12 = 0,

3y+ 12 = 0

y = 4

Therefore, the coordinates ofAare (0, 4).

(b)1(h) + 2(0)

1 + 2

= 3

h+ 0 = 9

h= 9

1(k) + 2( 4)

1 + 2

= 0

k 8 = 0 k= 8

(c)

O

A(0, 4)

B(3, 0)

y

h

x

Area of AOB

=12

(3) (4)

= 6 unit2

AB= 32+ ( 4)2

= 25 = 5 units

Let hbe the perpendicular distance from OtoAB.

Area of AOB= 6

12

(h)AB= 6

12

(h)(5) = 6

h=2 6

5

=125

units

15. (a) y 3x 5 = 0

y = 3x+ 5 .......................................

(2 + k)x+ 4y 6 = 0

4y= (2 + k)x+ 6

y= (2 + k)

4 x+

32

.............

Sinceand are parallel, therefore the gradients are the same.

3 = (2 + k)

4

2 + k = 12

k = 14

Substitutex= 1,y= tintoy 3x 5 = 0,

t 3(1) 5 = 0

t = 8


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25



(b) A(1, 8)

The line which is perpendicular toy 3x 5 = 0

has gradient of 13

.


y 8 = 13

(x 1)

y = 13x+

13

+ 8

y = 13x+

253

(c) (2 + k)x+ 4y 6 = 0

[2 + (14)]x+ 4y 6 = 0

12x+ 4y 6 = 0

6x+ 2y 3 = 0 ...............................

y = 13x+

253

.............

Substitute into ,

6x+ 2 13x+253 3 = 0

6x23x+

503

3 = 0

3 6x 23x+503

3 = 0 18x 2x+ 50 9 = 0

20x+ 41 = 0

x =4120

Substitutex=4120

into ,

y= 13

4120+

253

= 4160

+253

=15320

Therefore, the point of intersection is

4120 ,15320.

16. (a) Let P(x,y)

PB= 2PA (x 4)2+ (y 1)2 = 2(x 1)2+ (y 3)2

Square both sides,

(x 4)2+ (y 1)2= 4[(x 1)2+ (y 3)2]

x2 8x+ 16 +y2 2y+ 1

= 4(x2 2x+ 1 +y2 6y+ 9)

= 4x2 8x+ 4 + 4y2 24y+ 36

x2+ y2 8x 2y+ 17 = 4x2+ 4y2 8x 24y+ 40

4x2+ 4y2 8x 24y+ 40 x2 y2+ 8x+ 2y 17 = 0

3x2+ 3y2 22y+ 23 = 0

(b) Wheny= 0,

3x2+ 23 = 0

x2 = 233

x = 233 Since xdoes not have real values, therefore thelocus does not intersect the x-axis.

Whenx= 0,

3y2 22y+ 23 = 0

y =(22) (22)2 4(3)(23)

2(3)

=22 208

6

= 1.263, 6.070

Therefore, the locus intersects the y-axis at two

points.

17. (a) Gradient of CD= Gradient ofAB

52t t

3 0

=6 0

5 2

3t

2

3= 2

3t

2

= 6

t= 6 23

= 4

The equation ofADisx2

+y4

= 1.

(b)

A(2, 0)

B(5, 6)

E(x, y)1

3

3x+ 1 21 + 3

= 5

3x+ 2 = 20

3x = 18

x = 6

3y+ 1(0)

1 + 3

= 6

3y = 24

y = 8

Therefore, the coordinates ofEare (6, 8).


26/26


18. (a)

Q(2, 3) R(6, 3)

P(x, y)

Gradient of PQGradient of PR= 1

y 3x 2

y 3x 6

= 1 (y 3)2= 1(x 2)(x 6)

y2 6y+ 9 = (x2 8x+ 12)

= x2+ 8x 12

y2 6y+ 9 +x2 8x+ 12 = 0

x2+ y2 8x 6y+ 21 = 0

(b) x2+ y2 8x 6y+ 21 = 0 ........................ x= 2y

y=12x.....................................................

Substitute into ,

x2+ 12x2

8x 6 12x+ 21 = 0 x2+

14x2 8x 3x+ 21 = 0

54x2 11x+ 21 = 0

4, 5x2 44x+ 84 = 0

(x 6)(5x 14) = 0 5x 14 = 0

x=145

From ,y=12

145

=1410

=75

Therefore, the coordinates of Pare (145

,75

).

19. (a)

A(1, 3)B(4, 3)

C(4, 6)

D(p, q)1

2

p=1( 4) + 2(4)

1 + 2

=43

q=1(3) + 2(6)

1 + 2

= 5

The coordinates ofDare (43

, 5).

(b) Area of ABC

= 12

1 4 4 1

3 3 6 3 =

12

|[(3 24 + 12) (12 + 12 6)]|

=12|9|

=92

unit2

(c) Area of ADC=13

Area of ABC

=13

92

= 32

unit2

20. (a) P(1, 3), Q(5, 9),R(2, 12), S(x,y).

Midpoint of PR= Midpoint of QS

1 + 22

,3 + 12

2

= x+ 52

,y+ 9

2

1 + 2

2=

x+ 5

2 and

3 + 12

2=

y+ 9

2 x+ 5 = 1 y+ 9 = 15

x= 4 y= 6

(b) Area of PQRS

=12

1 5 2 4 1

3 9 12 6 3 =

12|[(9 + 60 + 12 12) (15 + 18 48 6)] |

=12|[51 (21)]|

=12

(51 + 21)

=12

(72)

= 36 unit2

(c) Gradient of PR=12 3

2 (1)

=93

= 3

The equation of PRis

y 3 = 3(x+ 1)

y 3 = 3x+ 3

y= 3x+ 6

Givenx 6

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