Waja 2009: Additional Mathematics Form Four

WAJA 2009

ADDITIONAL MATHEMATICS FORM FOUR

( Student’s Copy )

Name: ___________________________ Class : ___________________________

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

Learning Objective 6.1 Find distance between two points. Learning Outcome 6.1.1 Find the distance between two points using formula. 1. The formula of distance between two points is actually derived from _______________ theorem. 2.By referring to the diagram below, tick the correct equation : Equation (a) r 2  t 2  s 2 (b) t 2  s 2  r 2 (c) s 2  r 2  t 2 (d) r 2  t 2  s 2 (e) t 2  r 2  s 2 (f) s 2  r 2  t 2 (g) t 2  s 2  r 2 (h) s 2  t 2  r 2

t s r

 if correct

3.Complete the equation in the boxes : z

y2 

z2 

y x

x2 

4. Complete the following calculation : (i) A

Pythagoras Theorem : AC 2  AB 2  BC 2

5 cm B

12 cm

AB =

cm

BC =

cm

AC

C

+ = =

2

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

(ii)

Chapter

y PR 2  PQ 2  RQ 2

R (7 , 8)

PR 

8 RQ =

PQ 2  RQ 2

PR 

−



=

= 2

Q

P (3 , 2)

0

units.

x

7

3 PQ =

−

= (iii)

y

C

y2

BC =

− AC 2  AB 2  BC 2

y1

B

A

x2

x1

0

AB =

−

AC 

(

)2  (

x

Distance =

5.Find the distance between two points.2 Example : (i) Find the distance between C(-3 , 5) and D (2 , -3). Formula : Distance = ( x 2  x1 ) 2  ( y 2  y1 ) 2 Substitude : CD =

  3  2  2   5  [3] 2

Solve :



( 5) 2  (8) 2



25  64



89

 9.434 units

3

)2

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

(ii) Complete the calculation below : if the 2,a) is 7.

Formula : CD =

( x 2  x1 ) 2  ( y 2  y1 ) 2

Find the value of a distance between M (3,6) to N (-

Substitude :

 Solve :

2   6  a 2

7

) 2  (6  a ) 2  7

(

  2   6  a 2  7 2 (6  a ) 2 

(6  a )  

24

(6  a )   4.899

6  a  4.899 6  4.899  a a  1.101

6  a  4.899 6  4.899  a a  10.899

(a) Find the distance between P(2 , 5) and Q(-6 , 3). (b) Find the distance between S(-2, 4) and T(4 , 3). (c) Find the possible values of p if the distance between the two points A(1 , 2) and B(p , 14) is 13. (d) Find the possible values of m if the distance between the two points P(2 , m) And Q(-1 , 2) is 5. 6. Complete the table below : Situation

Implication

(a)

The distance from point M to point N is 5 cm

(b)

The distance of point D from origin is 7 cm.

(c)

Point A and point B is equal distant from point C

(d)

The distance from point R to point S is twice the distant from point L to point M .

(e)

Triangle PQR is an equilateral triangle.

(f)

The distance of point R from origin is three times of its distant from point S.

4

MN = 5 cm

PQ =

=

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

7. Locate the mistake done in the solution in column A and solve the problem correctly in column B. A (Wrong solution) (a) Given that the distance between points A (2, y) and B (1, 3) is 10 , find the values of y . Solution : AB = 10

B (Correct solution)

(12  2 2 )  (3 2  y 2 )  10

(1  4)  (9  y 2 )  10  3  9  y 2  10  0  y2  4  0 y2  4 y

4

y  2

(b) Find the distance between point L(9, 7) and point M(-3, 2). Solution :

ML= (9  [3]) 2  (7  2) 2 = (12) 2  (5) 2 = 144  25 = 119 = 10.91

(d) If point P(3 , k) is equidistant from points R(3 , -4) and S(-3 , 4), find the value of k. Solution : PR = RS (3  3)  ( k  [4]) 2  2

(3  [ 3]) 2  ( 4  4) 2

0 2  (k  4) 2  6 2  (8) 2 (k  4) 2  36  64 ( k  4)   100  10

k  4  10 k 6

k  4  4 k  8

5

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

A (Wrong solution) (d) Given the points A(3 , p), B(2 , 3) and C(1 , -2), find the possible values of p if the distance of point C to point A is twice its distance from point B. Solution : CA = 2CB (3  1) 2  ( p  [ 2]) 2  2 ( 2  1) 2  (3  [2]) 2 2 2  ( p  2) 2  2 12  5 2

4  ( p  2) 2  21  25

( p  2) 2  52  4 ( p  2)   48  6.928

p  2  6.928 p  4.928

p  2  6.928 p   8.928

6

Chapter

B (Correct solution)

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

Learning Objective : 6.2 Understand the concept of division of a line segment Learning Outcome : 6.2.1 Find the midpoint of two given points 1. In the following cases, mark the midpoint of the line segment below as M and state its coordinates : y K (a) y (b) 4

4

G 2

0

(c)

H

2

2 J

x

4

y

0

(d)

P

2

4

x

y R

4

4

2

2 L

L 0

2

x

4

0

2

4

6

x

2. Based on the diagram below, point M is the midpoint of the line segment AB, choose the correct coordinate for the midpoint M.

 x 2  x1 y 2  y1  , y  2 2   B

( x2 , y 2 )

 x 2 x1 y 2 yM1  ,   2 2   A ( x1 , y1 )

 x 2  x1 y 2  y1  , 0  2 2  

x

7

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

3. Match the diagram below :

Segment

Midpoint

A (4, -5) and B (2, -3 )

(-5.5 , 3)

C (-1, 5) and D (-7, -8)

(3a , 2.5b)

E (-4, 8) and F (-7, -2)

(3 , -4)

G(3a, 7b) and H(0, 6b)

(1.5a , 6.5b)

I(-4a, -3b) and (10a, 8b)

(-4 , -1.5)

4.The points P and Q are (a, 2) and (-2, b) respectively. The midpoint of PQ is (3, 4). Determine the value of a and the value of b. Sketch the segment :

Q

 P Apply formula :

 x 2  x1 y 2  y1  ,  2 2  

Midpoint = 

 2 

(3 , 4 ) =  

 2

,

  

Compare :  a  2 2  b ,  2 2  

(3 , 4 ) =  

Solve :

2 6a2

a



2 8 2b b

8

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

y

5. Find the coordinate of D if CM = MD.

D M (8,10) C (2,4) 0

Sketch the segment : * Let D ( x1 , y1 )  M (8 , 10)

x

D ( x1 , y1 )

C (2 , 4) Apply formula :  x 2  x1 y 2  y1  ,  2 2  

Midpoint = 

 2 

(8 , 10 ) =  

 2

,

  

Compare x-coordinates and y-coordinates : 

(8 , 10 ) =  



2

2

,



 

2 

2

The final answer must be in the form of D(x,y)

Solve :

9

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

6. P(0, 1) , Q(-8, 5) , R(-6, 9) and S are the vertices of a parallelogram. Find the coordinates of point S. Sketch : * Let S ( x1 , y1 ) R (-6 , 9)

Q (-8 , 5)

For square, rectangle, rhombus and parallelogram, the diagonals bisect each other. B A

S ()

P (0 , 1)

C

Apply formula :

D

 x 2  x1 y 2  y1  ,  2 2  

Midpoint = 

Midpoint of PR = Midpoint of QS   

 2

,

 2 







  

 2

,

 2

Compare :

Solve :

10







Midpoint = Midpoint of AC of BD

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

Learning Outcomes : 6.2.2 Find coordinates of a point that divides a line according to a given ratio m : n. 1. Complete the table below : Situation

Ratio

(a) C 3 2

AB : BC = 2 : 3

B

AB : AC = 2 : 5

A (b)

1

R

3

:

PQ : QR =

Q

:

PQ : PR =

P (c) K

KL : LM = 4 : 3 L

KL : KM =

:

RS : ST =

:

RS : RT = 2

: 7

FG : GH =

:

M (d) T

2

S

R (e) F

m G

n FG : FH = H

11

:

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

2. Tick the correct diagram base on the description given. Situation

Diagram I

(a) 3

P divides AB internally in the ratio 3 : 5

Diagram II B

P

5

N

1

L divides MN internally in the ratio 2 : 1

M G K

4

3 3

3

B

H

C

B

2 A

K 1

4

C

5 2

K

H

A (e) Three points K, L and M are on a straight line such that the length of KM is 3 times the length of KL.

N

L

1

G

(d) A, B and C are three points on a straight line such that AB : AC = 2 : 5 .

2

L

2 M

K divides GH internally in the ratio of 3 : 4 .



A

(b)

B

P

3

A

(c)

5

K L

1 2

12

M

L 3

M

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

3. A and B are the points (8 , 4) and (-2 , -6) respectively. Point P divides line AB in the ratio 3 : 2 . Find the coordinates of P. Sketch : Let P (x , y)

B (-2, -6) B

P (x , y)

n P

A (8 , 4)

m A

Formula :  mx 2  nx1 my 2  ny1  ,  mn mn    3( )  2( ) 3( )  2( )  ,  (x , y ) =  3 2 3 2  

P (x , y ) = 



(x , y ) =  

Compare : (x , y ) = 

5

,

Coordinates of P =

 

,

5 



Coordinate of P is 4. A line segment has end points at P (-4, -8) and S (a , 6). Given that point R (2, b) is a point on the straight line PS such that 3PR= 4PS. Find the value of a and b . Sketch :

S (a, 6) 3PR = 4RS

R (2 ,b)

P (-4 , -8)

13

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

Formula :  mx 2  nx1 my 2  ny1  ,  mn mn  

R (2 , b ) = 

4( )  3 ( )   4( )  3( ) ,  (2 , b ) = 43 43   

(2 , b ) =  

7

,

 

7 

Compare : 2=

b=

Solve :

5. Find the ratio m : n if a certain point divides a line segment internally in the ratio m : n and the coordinates of the points are Sketch : given. Example : If point P(3 , 4) internally divides the line segment that connects the points of m :

Formula :  mx 2  nx1 my 2  ny1  ,  mn m  ,n7)   B(6 m( 7 )  n ( 2 )   m( 6 )  n( 1 ) ,  (3 , 4 ) =   mm  n P (3, 4) m  n

P (3 , 4 ) = 

6m  n 7 m  2n  ,   n2) mn   Am(1, 

(3 , 4 ) = 

Compare (either one): 3

6m  n mn

4

Either one

7 m  2n mn

Solve :

6m  n mn 3m  3n  6m  n 3m  6m  n  3n  3m  2n m 2  n 3 m:n  2:3 3

14

A(1 , 2) and B(6 , 7) in the ratio n , find the ratio m : n .

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

Chapter

Sketch : (5 , 3

(a) R joining in the

1 ) divides the line 2

A(2 , 8) and B(6 , 2) internally ratio m : n . Find the ratio m : n . A(1 , 2) Formula : R (5 , 3

1  mx 2  nx1 my 2  ny1  ,  ) = 2 mn mn  

 m( )  n(  1 mn (5 , 3 ) = 2

(5 , 3

1  ) = 2 

mn

,

)

,

m( )  n ( )   mn   

mn 

Compare : 7 = 2

5=

Solve :

m  n

m:n 

15

:

WAJA 2009 – ADDITIONAL MATHEMATICS (FORM 4) 6: Coordinate Geometry

16

Chapter