6.coordinate geometry

8/22/2019 6.Coordinate Geometry

1/26

1

Additional Mathematics SPM Chapter 6

Penerbitan Pelangi Sdn. Bhd.

6Coordinate Geometry

1. (a) AB = (1 4)2 + (2 6)2

= 9 + 16= 5 units

(b) PQ = (1 + 3)2 + (2 4)2

= 16 + 36= 52 units

(c) RS = (3 + 1)2 + (4 2)2

= 4 + 4= 8 units

(d) UW= (1 4)2 + (0 + 5)2

= 25 + 25= 50 units

(e) CD = (2 2)2 + (0 5)2

= 0 + 25= 5 units

(f) EF = (0 3)2 + (4 + 4)2

= 9 + 0= 3 units

(g) GH= 12 42

+ (5 + 0.3)2

= 72 2

+ ( 4.7)2

= 34.34= 5.86 units

2. AB = 10(2 1)2 + (y 3)2 = 10

1 + (y 3)2 = 10

(y 3)2 = 9

y 3 = 3

y = 3 + 3

y = 0, 6

3. PQ = 16(a 2)2 + [(a + 1) + 1]2 = 16

(a 2)2 + (a + 2)2 = 16

a2 4a + 4 + a2 + 4a + 4 = 16

2a2

+ 8 = 16 a2 = 4

a = 2

4. AC =AB

(p + 1)2 + (2 2)2 = (3 + 1)2 + (5 2)2(p + 1)2 + 16 = 16 + 9

(p + 1)2 = 9

p + 1 = 3

p = 3 1

= 4, 2

Since C is in the quadrant IV, therefore p = 2.

5. (a) Midpoint ofAB = 1 + 52

, 3 + 72

= (3, 5)

(b) Midpoint ofCD = 1 + 92

, 5 12

= (4, 2)

(c) Midpoint ofEF = 2 42

, 3 + 52

= (3, 4)

(d) Midpoint ofGH=

8 2

2

, 10 6

2 = (5, 8)

(e) Midpoint ofIJ = 0 + 82

, 4 22

= (4, 1)

(f) Midpoint ofKL = 8 1

3

2

, 0.6 + 62

= 25

6, 3.3


2/26

2



6. Midpoint ofAB = (3, 4)

1 + 52

, t+ 22

= (3, 4) t+ 2

2= 4

t+ 2 = 8

t= 6

7. Midpoint ofPQ = (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 ofPR.

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 ofC are (9, 6).

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

2

+ 4

2

= 0.4, 94 Hence,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

= 10

3, 6

The coordinates ofP are 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 ofP are 3, 10

3.

(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 = 3

5,

13

5


,13

5

.(c)

2

1 12

B(6, 3)

P(x, y)

A(, 4)

(x,y) =

1(6) + 2 12

1 + 2,

1(3) + 2(4)

1 + 2

= 53

,11

3


,11

3 .

(d)21

21

1B(, 0)

P(x, y)

A(3, 2)

(x,y) =

1

1

2 +1

2 (3)1

2

+ 1

, 1(0) +

1

2 (2)1

2

+ 1 = 2

3,

23


, 2

3

.


3/26

3



13.QR

RS

=1

3

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, 133

The coordinates ofQ are 0, 13

3.

14. PS =1

3RS

PS

RS

=1

3

PS :PR = 1 : 2

2

21

1

R(, 4)

P(x, y)

S(0, 8)

(x,y) = 2(0) + 11

2

1 + 2,

2(8) + 1(4)

1 + 2 =

16 , 4


, 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 ofB are 72

, 2.

16. (a) Area ofABC =1

2

0 1 3 0

4 2 5 4=

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

=1

217 10

=7

2

unit2

(b) Area ofABC

=1

2

1 4 5 1

3 2 6 3=

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

=1

241 + 4

=45

2

unit2

(c) Area ofABC

=1

2

0 4 2 0

1 3 5 1=

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

=

1

2 18 2

=1

220

=1

2

(20)

= 10 unit2

(d) Area ofABC

=1

2

1 2 3 1

2 4 6 2=

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

= 0 unit2

17. (a) Area ofABCD

=1

2

1 4 3 2 1

2 5 6 3 2=

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

=1

242 38

= 2 unit2

(b) Area ofABCD

=

1

2

1 2 3 2 1

1 1 5 7 1

=1

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

=1

234 + 22

= 28 unit2


4/26

4



(c) Area ofABCD

=1

2

0 1 2 3 0

3 4 1 1 3=

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

=1

26 + 8

=1

22

= 1 unit2

(d) Area ofABCD

=1

2

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 ofPQR =1

2

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, Q and

R are collinear.

19. Area ofOBC = 132

1

2

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 ofPQRS

=1

2

0 1 2 3 0

1 4 7 10 1

=1

2(0 + 7 + 20 + 3) (1 + 8 + 21 + 0)

=1

230 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 = 1

2

;y-intercept = 1.

(b) 2xy + 3 = 0

Whenx= 0, 0 y + 3 = 0

y = 3

Wheny = 0, 2x 0 + 3 = 0

x= 3

2

x-intercept = 3

2

;y-intercept = 3.

(c)x

2 +y

3 = 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)

= 6

4

= 3

2

(d) Gradient =3 0

4 (5)

= 3

23. (a) Gradient = y-intercept

x-intercept

= 3

2

(b) Gradient = 2

3=

23

(c) Gradient = 42

3

= 4 32

= 6


5/26

5



(d)x

2

y

5

= 2

x

4

y

10

= 1

Gradient = 10

4

=5

2

(e)x

4

+y

3

=1

2

2 x4

+y

3

= 2 12

x2

+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 =1

4

(x 2)

y =1

4x

12

6

y =1

4x

13

2

25. (a) The equation of lineAB is

y 1x 2

=4 1

3 2

= 3

y 1 = 3(x 2)

= 3x 6

3xy 5 = 0

(b) The equation of lineAB is

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 lineAB is

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

x

3

+y

4

= 1

(b)x

3

+y

1

= 1

x3

y = 1

(c)x

1

+y

2

= 1

xy

2

= 1

(d)x

12

+y

4

= 1

2xy

4

= 1

27. (a) y = 3x+ 1

Gradient, m = 3

y-intercept = 1Wheny = 0, 0 = 3x+ 1

x= 1

3

x-intercept = 1

3

(b) 2y = 4x 3

y = 2x3

2

Gradient, m = 2

y-intercept = 3

2

Wheny = 0, 2x= 3

2

x= 34

x-intercept = 3

4

(c) 2x+y = 5

y = 2x+ 5

Gradient , m = 2

y-intercept = 5

Wheny = 0, 2x= 5

x=5

2

x-intercept =5

2


6/26

6



(d) 2y 1

2x+ 5 = 0

2y =1

2x 5

y =1

4x

52

Gradient, m =1

4

y-intercept = 5

2

Wheny = 0,1

4x=

52

x= 10

x-intercept = 10

(e)x

2

+y

3

= 1

Gradient , m =

3

2x-intercept = 2

y-intercept = 3

(f)1

2x

13y + 4 = 0

1

2x

13y = 4

12x

4

13y

4

= 4 4

x

8

+y

12

= 1

Gradient, m = 12

8=

32

x-intercept = 8

y-intercept = 12

28. (a) 2y = 3x 1

3x 2y 1 = 0

(b)x

2

=y

3

+ 1

6 x2

= 6 y3

+ 13x= 2y + 6

3x 2y 6 = 0

(c)x+ 1

3=y

4

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

x

2

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

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

, 8y = 7

y = 78

Substitutey = 7

8

into ,

x+ 2 78

= 1 x= 1 +

7

4

=11

4

Point of intersection = 114

, 7

8

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

6x 2y = 1 ................................

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

y =16

11

Substitutey =16

11

into ,

2x+ 3 1611

= 52x= 5

4811

=7

11

x=7

22

Point of intersection 722 , 1611 30. (a) y = 2x 1

Gradient = 2

2y = 4x+ 3

y = 2x+3

2

Gradient = 2

Hence, the two lines are parallel.


7/26

7



(b) 3xy + 4 = 0

y = 3x+ 4

Gradient = 3

3x+y 5 = 0 y = 3x+ 5

Gradient = 3

Hence, the two lines are not parallel.

(c)x

2

+y

3

= 1

Gradient = 3

2

2y = 3x 5

y = 3

2x

52

Gradient = 3

2

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

2

x 5

Gradient =k

2


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

3

x+4

3

Gradient =k

3


k

3

= 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


y 3 = 2(x+ 1)

y = 2x+ 2 + 3

y = 2x+ 5

(c) 4xy + 1 = 0

y = 4x+ 1

Gradient = 4


y + 2 = 4(x 0)

y = 4x 2

(d)x2

y

6

= 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.


8/26

8



(b) 2y = 6x+ 5

y = 3x+52

Gradient = 3

y =13x 4

Gradient =13

m1m

2= (3) 1

3

= 1


(c) x+ 2y = 5

2y = x+ 5

y = 12x+

52

Gradient = 1

2

2y 4x= 7

2y = 4x+ 7

y = 2x+72

Gradient = 2

m1m

2= 1

2(2)

= 1


(d) xy = 8

y =x 8

Gradient = 12x+y = 1

y = 2x+ 1

Gradient = 2

m1m

2= (1)(2)

= 2

The two lines are not perpendicular.

(e)x2

y

4

= 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 = 2

k

y =16x 1

Gradient =16

m1m

2= 1

2k 16 = 1

13k

= 1

3k= 1

k=13

(c) 2y + 4kx= 3

2y = 4kx+ 3

y = 2kx+32

Gradient = 2k

x2

+y

6

= 1

Gradient = 6

2

= 3

m1m

2= 1

(2k)(3) = 1

6k= 1

k= 16

(d)12kx+ 2y = 5

2y =

1

2 kx+ 5

y = 14kx+

52

Gradient = 1

4k

4x+ 3y = 6

3y = 4x+ 6

y = 43x+ 2

Gradient = 4

3


9/26

9



m1m

2= 1

14k 4

3 = 1

k3

= 1

k= 3

35. (a) y = 4x 1

Gradient = 4

The equation of the perpendicular line is

y 3 = 1

4

(x 1)

y = 1

4x+

14

+ 3

y = 1

4x+

134

(b) y = 12x+ 4

Gradient = 12


y 2 = 2(x+ 1)

y = 2x+ 2 + 2

y = 2x+ 4

(c) 2xy = 2

y = 2x 2

Gradient = 2


y + 3 =

1

2 (x 0)

y = 12x 3

(d)x3

+y

4

= 1

Gradient = 43


y + 2 =34

(x+ 1)

y =34x+

34

2

y =34x

54

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

= , 2x 1 = 4x+ 32x= 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. 2xy = 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

=32

Gradient ofPQ = 23

The equation of line PQ is

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 is x+y 3 = 0.


10/26

10



(b)PAPB

=12

PB = 2PA

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

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

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

3x2 + 3y2 12x 10y + 7 = 0

Hence, the equation of locus is

3x2 + 3y2 12x 10y + 7 = 0.

(c) PAPB

=2

3

3PA = 2PB

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

9(x2

+ 2x+ 1 +y2

8y + 16)= 4(x2 + 4x+ 4 +y2 + 6y + 9)

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

= 4x2 + 16x+ 16 + 4y2 + 24y + 36

5x2 + 5y2 + 2x 96y + 101 = 0

The equation of locus is

5x2 + 5y2 + 2x 96y + 101 = 0.

41. Substitutey = 0 into x2 +y2 = 4,

x2 = 4

x= 2

The points of intersection are (2, 0) and (2, 0).

42. PA =PO (x 1)2 + (y 2)2 = x2 +y2

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

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

2x+ 4y 5 = 0

The equation of locusP is 2x+ 4y 5 = 0.

Whenx= 0, 4y 5 = 0

y =54

y-intercept =54

Wheny = 0, 2x 5 = 0

x=5

2x-intercept =

52

43. x2 +y2 200 = 0 ................................. yx= 0 .................................From ,y =x....................................

Substitute into ,x2 +x2 200 = 0

2x2 = 200

x2 = 100

x= 10

Substitutex= 10 into , y = 10

The points of intersection are (10, 10) and (10, 10).

44.

0

y

x

5 units

5 units

The equations of locus are y = 5 andy = 5.

45.

0

y

xA(1, 2)

P(x, y)

PA =y

(x 1)2 + (y 2)2 =y(x 1)2 + (y 2)2 =y2

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

x2 2x 4y + 5 = 0

The equation of the locus ofP isx2 2x 4y + 5 = 0.

1.

3

2

P(2r, 5s)

R(t, 2t)

Q(r, s)

(r, s) = 2(2r) + 3(t)3 + 2

,2(5s) + 3(2t)

3 + 2

= 4r+ 3t5

,10s + 6t

5

r= 4r+ 3t

55r= 4r+ 3t

r= 3t.............................................

and s =10s + 6t

5

5s = 10s + 6t

6t= 5s

t= 56s ....................................

Substitute into ,

r= 3 56s

r= 52s


11/26

11



2. y = mxc

Gradient = m

y = (3 c)x+ m

Gradient = 3 c

m1m

2= 1

(m)(3 c) = 1

3 c = 1m

c = 3 +1m

3. 2x+ 4y 1 = 0

4y = 2x+ 1

y = 12x+

14

Gradient = 12

y6

x3

= 1

y6

+x

3

= 1

Gradient = 6

3

= 2

m1m

2= 1

2(2)

= 1

The two straight lines are perpendicular to each

other.

4.

x

4 +

y

5 = 1

Gradient = 54 =

54

Coordinates ofP = (4, 0)


y 0 = 45

(x+ 4)

y = 45x

165

5. y = 2x+ 1

Gradient ofQR = 2

Gradient ofPQ =12

The equation of linePQ is

y =12x+ 2 ...................................

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

= ,12x+ 2 = 2x+ 1

12x+ 2x= 1 2

52x= 1

x= 2

5

Substitutex= 25

into ,

y = 2 25 + 1=

95

The coordinates ofQ are 25

,95.

6.12

x1x

2x

3x

1 y1y

2y

3y

1

= 8

12

1 4 2 1

3 h 0 3

= 8

12

[(h + 0 + 6) (12 + 2h + 0)] = 8

h + 6 12 2h = 16

3h 6 = 16

3h = 16 + 6

3h = 16 + 6 , 3h = 16 + 6

3h = 22 , 3h = 10

h = 223

h =103

=103

7. (a) The equation ofPQ is x

4+y

8= 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 ofS are (1, 6).

(c) x4

+y8

= 1

Gradient ofPQ =

8

4= 2

Gradient ofRS = 12

Let the coordinates ofR be (x1, 0).

0 6

x

1 (1)

= 12

6

x

1 (1)

= 12

x1

+ 1 = 12

x1

= 11

Hence, thex-intercept ofRS is 11.


12/26

12



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 P is

x2 +y2 6x 4y = 0.

(ii) Substitute D(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 ofCE =32

Let the coordinates ofE be (0,y1).

y

1 0

0 6

=32

y1

=32

(6)

y1

= 9

Area ofCOE =12

6 9

= 27 unit

2

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

2y = x+ 6

y = 12x+ 3

Gradient ofPQ = 12

Gradient ofRQ = 2

The equation of line RQ is

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 ofQ are 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

16

5

=3 + 2x

5

and7

5

=2y 9

52y 9 = 7

y = 8

3 + 2x = 16

x =13

2

The coordinates ofS are 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 point Mis

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

10. (a) Area ofABC

=12

0 2 2 0

3 1 4 3=

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

=1214 + 4

= 9 unit2


13/26

13



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

1 + 3

= 1,11

4

(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. E is 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 4

3,y3

x 4

3= 0 and

y3

= 4

x= 4 y = 12


5. Let the coordinates ofD be (0,y).

Gradient ofCD = Gradient ofAC

y 60 3

=6 1

3 (2)

y 6 = 3(1) y = 3

Area ofBCD =12

0 5 3 0

3 2 6 3=

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

=1239 21

= 9 unit2

6. Area of quadrilateralPQRS

=

1

2

0 5 2 1 0

3 2 6 1 3= 12(0 + 30 + 2 + 3) (15 4 6 + 0)

=1235 + 25

= 30 unit2

7. Area ofABC = 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 line ABCD is

y 5 = 1(x 3)

y =x 3 + 5

y =x+ 2


14/26

14



(b) y-intercept = 2

Wheny = 0, 0 = x+ 2

x= 2

x-intercept = 2

9. (a) Gradient ofRQ = 2

Gradient ofPQ=12

The equation ofPQ is

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 ofRQ is12

.

10. (a) 2xy = 4

y = 2x 4

Gradient ofCD = 2

Gradient ofAB = 2

The equation of lineAB is

y 5 = 2(x 2)

y = 2x 4 + 5

y = 2x+ 1

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

Substitute into ,2x (x 2) = 4

2x+x+ 2 = 4

3x= 2

x=23

Substitutex=23

into ,

y = 23 2

=

8

3

The coordinates ofD are 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 + 16

5x2 + 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

(b) Gradient ofOA =3 0

1 0

= 3

Gradient of tangent atA = 1

3

The equation of the tangent at A is

y 3 = 13

(x 1)

y = 1

3

x+1

3

+ 3

y = 1

3

x+4

3

14. (a) Let the coordinates ofC be (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 line AD is

y 0 = 1(x 4)

y = x+ 4


15/26

15



(c) Let the point of intersection ofBC at the y-axis

beE(0,y).

Gradient ofBD = Gradient ofBE

1 = y (2)

0 2

2 =y + 2

y = 4

They-intercept of lineBC is 4.

15. (a) Area ofPQR

=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 linePQ and they-axis be

S(0,y1).

Gradient ofPS = Gradient ofPQ

y

1 5

0 3

=5 3

3 (2)

y1

5 =25

(3)

y1

= 6

5

+ 5

=19

5

They-intercept of linePQ is19

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 lineBC and they-axis be

E(0,y).

Gradient ofBE = Gradient ofBC

y 40 3

=4 (8)

3 (1)

y 4 = 3 124

= 9

y = 5

They-intercept of lineBC is 5.

(b) Gradient ofAD = Gradient ofBC

=4 (8)

3 (1)

= 3

The equation of line AD is

y 6 = 3(x+ 3)

= 3x+ 9

y = 3x+ 15

(c) Let the coordinates ofD be (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 ofD are (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 ofCD

=0 (3)

2 (1)

= 1

The equation of line BC is

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 ofB are (8, 6).

(b)

3

2C(1, 3)

B(8, 6)

E(x, y)


16/26

16



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

,3y + 2(6)

2 + 3

=

3x+ 16

5

,3y + 12

5

3x+ 16

5

= 1 and3y + 12

5

= 3

3x= 21 3y = 27

x= 7 y = 9

The coordinates ofE are (7, 9).

(c) x 2y + 4 = 0

Whenx= 0, 2y + 4 = 0

y = 2

F(0, 2)

Area ofBCF

=

1

2

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 ofCD = 2

The equation of line CD is

y + 3 = 2(x 1)

= 2x+ 2

y = 2x 1

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

k= 2(2) + 6

k= 2

Gradient ofCE = Gradient ofBC

0 (3)

p 1

=2 (3)

2 1

3 = 5(p 1)

p 1 =35

p =85

(c)

n

mE(, 0)

58

C(1, 3)

B(2, 2)

Usey-coordinate,

(3)n + 2m

m + n

= 0

2m 3n = 02m = 3n

mn

=32

CE :EB = 3 : 2

(d) Area ofBOC =12

0 1 2 0

0 3 2 0=

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

=122 + 6

= 4 unit2

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

PAPB

=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 43 2

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

16

3

163

= 0

= RHS

Hence, the point 43

, 0 lies on the locus ofP.

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

3x2 + 4x= 0

x(3x+ 4) = 0

x= 0 or 3x+ 4 = 0

x=

4

3

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.


17/26

17



20. (a) Area ofABC

=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 fromB to line

AC.

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

= 81 + 1= 82 units

Area ofABC = 20

12 d82 = 20

d=40

82

= 4.417 units

(b)

P(1, 3)

Q(h, k)

Midpoint ofPQ = 1 + h2

,3 + k

2

Since the midpoint ofPQ lies on the perpendicularbisector, so we substitute x = 1 + h

2and

y = 3 + k2

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 = 35 x+16

5

Gradient of perpendicular bisector = 35

Gradient of linePQ =53


y + 3 =5

3

(x+ 1)

=5

3x+

53

y =53x

43

Substitutex= h,y = kinto the equation ofPQ,

k=53h

43

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

Substitute into ,

3h + 5 53 h 43 = 50

3h +25

3

h 20

3

= 50

33h + 253

h 20

3

= 3(50)9h + 25h 20 = 150

34h = 170

h = 5

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

k= 7

21. (a)PAPB

=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 + 12 x+ 12

4 12 x+ 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


18/26

18



Substitutex= 65

into ,

y =1

2

6

5 + 1

= 35

+ 1

=25

The coordinates ofD are 65 ,25 .

(d) Gradient ofAC =12

Gradient ofBD =

25

(2)

65

0

= 12

5 56 = 2

Gradient ofAC Gradient ofBD =12

(2)

= 1

Hence, linesAC andBD are 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 + q

1 + 3

= 3p4 ,q

4

3p

4

=x andq

4=y

p = 4x3 q = 4y

Substitutep =4x

3

and q = 4y into

q2 +p2 = 100,

(4y)2 + 4x3 2

= 100

16y2 +16

9x2 = 100

16

9x2 + 16y2 100 = 0

The equation of the locus of point R is

16

9x2 + 16y2 100 = 0.

(ii) Substitutey = 0 into16

9x2 + 16y2 100 = 0,

16

9x2 100 = 0

x2 = 100 916 x= 900

16

= 30

4

= 15

2

Thex-coordinate ofR is 15

2.

23. (a) Gradient ofPQ Gradient ofRQ = 1

5 21 4

t 2r 4

= 1

(1) t 2r 4

= 1 t 2 = r 4

t= r 2

(b) Area ofPQR

=12

1 r 4 1

5 t 2 5=

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

=12

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

=12

(3t 3r+ 18)

= 32t

32r+ 9

= 9 32

(r+ t)

(c) Given the area of rectanglePQRS = 30 unit2

Area ofPQR = 15 unit2

9 32

(r+ t) = 15

3

2

(r+ t) = 6

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

Substitute into ,r+ r 2 = 4

2r= 2

r= 1

Substitute r= 1 into ,t = 1 2

= 3

The coordinates ofR are (1, 3)


19/26

19



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

22 + t2 = 16 t2 = 12

t= 12

Based on the diagram, t= 12

Gradient ofOA =12 0

2 0

=12

2

=23

2

12 = 4 3= 4 3= 23

= 3Gradient of tangentAB is 1

3

Equation of tangentAB is

y 12 = 13

(x 2)

y = 1

3

x+2

3

+ 12

= 1

3

x+2

3

+ 23

2. LetP(x,y)

Gradient ofPQ = 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

Substitutex= 2 into ,y = 2

The coordinates ofP are (2, 2).

Area of trapeziumPQRS

=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 ofAB Gradient ofBC = 1

6 (2)

3 (1)

k 6h 3

= 1

2k 6

h 3 = 12(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


20/26

20



4. (a) Area ofABC = 4

Since there are two possible positions for point

C,

therefore 12

1 2 k 1

1 5 2k 1 = 4[(5 + 4kk) (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

or15

4

(b) Gradient ofAB Gradient 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 ofPQ = 4 + r2

,9 + t

2

(b)

02y+ x= 7

7

2

7

y

A

P(4, 9)

Bx

Gradient ofPQ Gradient ofAB = 1

t 9r 4

7

2

7

= 1 t 9

r 4 1

2 = 1

t 9

r 4

= 2

t 9 = 2(r 4)

= 2r 8 t= 2r+ 1

(c) Ifr= 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 ofE are (0, 6).

Wheny = 0,

y + 3x 6 = 0

0 + 3x 6 = 0

x= 2

The coordinates ofF are (2, 0).

LetB = (x,y)

SinceE is the midpoint ofBF,

thenx+ 2

2

= 0

x= 2

y + 0

2= 6

y = 12

Therefore, the coordinates of B are

(2, 12).


21/26

21



(ii) Area of quadrilateral OABE

=12 0 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 ofD are (34, 48).

(c) (i) mAC = mAB

y 0

0 + 14

=12 0

2 + 14

y14

=12

12

y = 14


Let the moving point beP(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 ofA are (2, 4).

(b) LetB(x,y)

1(x) + 3(0)

1 + 3

= 2

x= 8

1(y) + 3(0)

1 + 3

= 4

y = 16

The coordinates ofB are (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 = 12 x+ 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 ofP are

(2, 0).

The coordinates ofQ are (0, 19).

Gradient ofPQ =

0 (19)

2 0

=192


y 0 =192

(x 2)

y =192

x 19


22/26

22



(b) Gradient of linePS = 2

19

The equation of line PS is

y 0 = 219

(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 + 219

x 23219

= 0

19, 57x2 + 2x 232 = 0

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

2(57)

=2 52 900

114

= 11657

, 2

x= 2 is ignored

because it is

x-coordinate for

pointP.

Substitute x= 11657

into ,

y = 2

19

11657 +

419

=460

1083

The coordinates ofS are 11657

,460

1083

.

9. (a) (i) Gradient ofPR =8 6

6 8

= 1

Gradient ofAC = 1 SincePR//AC

The equation of line AC is

y 12 = 1(x 10)

= x+ 10

y = x+ 22

(ii) The perpendicular bisector ofBC isPR.

Gradient ofPR = 1

The equation of line PR is

y 6 = 1(x 8)

y 6 = x+ 8

y = x+ 14

(b) Area ofPQR

=12

6 10 8 6

8 12 6 8 =

12

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

= 8 unit2

Area ofABC = 22(8)

= 32 unit2

Area ofPQR : Area ofABC

= 8 : 32= 1 : 4

(c) LetB(x,y)

SinceP is the midpoint ofAB

x+ 82

= 6 andy + 14

2= 8

y = 2x= 4

Therefore, the coordinates ofB are (4, 2).

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 ofOB = 2 08 0

= 14

The equation ofOB isy = 14x.

11. (a) Wheny = 0,

y2 = 6x+ 9

02 = 6x+ 9

6x= 9

x= 96

= 32

The coordinates ofQ are (32

, 0).

Whenx= 0,

y2 = 6(0) + 9

y2 = 9

y = 3

The coordinates ofP are (0, 3).


23/26

23



The equation ofPQ is

y 3 =3 0

0

32

(x 0)

y 3 = 2x

y = 2x+ 3

(b) Gradient ofQS = 12

mPQ

mQS

= 1

The equation of line QS is

y 0 = 1

2 x+

32

y = 1

2x

34

(c) y = 1

2x

34

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

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

Substitute into ,

12 x34

2

= 6x+ 9

14x2 + 2 1

2x 3

4 + 3

4

2

= 6x+ 9

14x2 +

34x+

916

= 6x+ 9

16 14 x2 +

34x+

916 = 16(6x+ 9)

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

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

4x2 84x 135 = 0

(2x+ 3)(2x 45) = 02x 45 = 0

x =452

Substitutex=45

2into ,

y = 1

2 45

2 3

4

= 454

34

= 12

Therefore, the coordinates ofS are (452

, 12).

12. (a) SincePQRS is a parallelogram,Midpoint ofPR = Midpoint ofQS

h + 62

,2k 5

2

= 2h 12

,k+ 1 + 4

2

h + 6

2

=2h 1

2

and2k 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

PR and QS.

T(x,y) = Midpoint ofPR x=

6 + h

2

=6 + 7

2

=132

y =2k 5

2

=20 5

2

=152

Therefore, the point of intersection of diagonals

PR and QS is T(132

,152

).

(c) Gradient ofQR =11 + 5

14 6

=168

= 2

The equation of line passing through T and is

parallel to QR is

y 152

= 2x 132 = 2x 13

y = 2x 13 + 152

y = 2x11

2

13. (a) A = 1 + 32

,8 10

2

= (2, 1)

(b) Midpoint ofPR = Midpoint ofQS

1 + 32

,8 10

2

= h 42

,k+ 5

2

(2, 1) = h 4

2,k+ 5

2

h 42

= 2 and k+ 52

= 1

h 4 = 4 k+ 5 = 2

h = 8 k = 7

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

Gradient ofPS =8 5

1 ( 4)

=35

x= 32

is ignored

because it is

x-coordinate ofQ.


24/26

24



The equation of the line passing through A and

parallel toPS is

y (1) =3

5

(x 2)

y + 1 =35x

65

y =35x

65

1

y =35x

115

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 ofA are (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 ofAOB

=12

(3) (4)

= 6 unit2

AB = 32 + ( 4)2

= 25= 5 units

Let h be the perpendicular distance from O to

AB.

Area ofAOB = 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

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

Since and 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

(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+

25

3

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

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

12x+ 4y 6 = 0

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

y = 13x+

253

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


25/26

25



Substitute into ,

6x+ 2 13 x+253 3 = 0

6x 23x+ 50

3 3 = 0

3 6x 23 x+503

3 = 018x 2x+ 50 9 = 0

20x+ 41 = 0

x =4120

Substitutex=4120

into ,

y = 13 41

20 + 25

3

= 41

60

+25

3=

15320

Therefore, the point of intersection is

(4120

,153

20

).

16. (a) LetP(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 = 233Since xdoes not have real values, therefore the

locus 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 ofCD = Gradient ofAB

52tt

3 0

=6 0

5 2

3t

2

3= 2

3t

2

= 6

t= 6 23

= 4

The equation ofAD isx2

+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 = 63y = 24

y = 8

Therefore, the coordinates ofE are (6, 8).

18. (a)

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

P(x, y)

Gradient ofPQ Gradient ofPR = 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


26/26

26



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

x= 2y

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

Substitute into ,

x2 + 12x

2

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 ofP are (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 ofD are (43

, 5).

(b) Area ofABC

=12

1 4 4 1

3 3 6 3=

12

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

=12

|9|

=92

unit2

(c) Area ofADC =13

Area ofABC

=1

3

9

2=

32

unit2

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

Midpoint ofPR = Midpoint ofQS

1 + 22

,3 + 12

2

= x+ 52

,y + 9

2

1 + 2

2=x+ 5

2and

3 + 12

2=y + 9

2 x+ 5 = 1 y + 9 = 15

x= 4 y = 6

(b) Area ofPQRS

=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 ofPR =12 3

2 (1)

=93

= 3

The equation ofPR is

y 3 = 3(x+ 1)

y 3 = 3x+ 3

y = 3x+ 6

Givenx 6

6.coordinate geometry

Documents