6.2 Division of a Line Segment


6.2 Division of a Line Segment

(A) Midpoints of a Line Segment


Formula for the midpoint, of (xl, y1) and (x2, y2) is




Example 1:
Given B (m – 4, 3) is the midpoint of the straight line joining A(–1, n) and C (5, 8). Find the values of and n.

Solution:

B  is the midpoint of  A C ( m 4 ,   3 ) = ( 1 + 5 2 ,   n + 8 2 ) ( m 4 ,   3 ) = ( 2 ,   n + 8 2 ) m 4 = 2  and n + 8 2 = 3 m = 6 and    n + 8 = 6    n = 2


(B) Point that Internally Divides a Line Segment in the Ratio m : n



Formula for the point that lies on AB such that AP : PB = m : n is




Example 2:
The coordinate of R(2, –1) divide internally the line of AB with the ratio 3 : 2. If coordinate of is (–1, 2), find the coordinate of B.

Solution:


Let point  B = ( p ,   q ) ( 2 ( 1 ) + 3 p 3 + 2 ,   2 ( 2 ) + 3 q 3 + 2 ) = ( 2 , 1 ) ( 2 + 3 p 5 ,   4 + 3 q 5 ) = ( 2 , 1 ) Equating the  x -coordinates, 2 + 3 p 5 = 2 2 + 3 p = 10 3 p = 12 p = 4 Equating the  y -coordinates, 4 + 3 q 5 = 1 4 + 3 q = 5 3 q = 9 q = 3  The coordinates of point  B = ( 4 , 3 ) .

6.1 Distance between Two Points

6.1 Distance between Two Points
(x1, y1) and (x2, y2) are two points on a coordinate plane as shown below. BC is parallel to the x-axis and AB is parallel to the y-axis.  Hence  ∆ ABC = 90°.


Distance between Point A and C =  



Example:
Find the distance between the points P (2, 2) and Q (–4, –5).

Solution:
Let P (2, 2) = (x1, y1 ) and Q (–4, –5) = (x2, y2 ).

Distance of  P Q = ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 = ( 4 2 ) 2 + ( 5 ( 2 ) ) 2 = 36 + 9 = 45 = 9 × 5 = 3 5  or  ( 6.71 )



Long Questions (Question 3 – 4)


5.6.2 Indices and Logarithms, SPM Practice (Long Questions)

Question 3:
Given that = 3r and q = 3t, express the following in terms of and/ or t.
(a)  (a)  log 3 p q 2 27 ,
(b)  log9p – log27 q.

Solution:
(a)
Given p = 3r, log3 p = r
q= 3t, log3 q =t

log 3 p q 2 27
= log3 pq2 – log327
= log3 p + log3 q2 – log3 33
= r + 2 log3 q – 3 log3 3
= r + 2 log3 q – 3(1)
= r + 2t – 3

(b)
log9 p– log27 q
= log 3 p log 3 9 log 3 q log 3 27 = r log 3 3 2 t log 3 3 3 = r 2 log 3 3 t 3 log 3 3 = r 2 t 3   



Question 4:
(a)  Simplify:
log2(2x + 1) – 5 log4 x2 + 4 log2 x
(b)  Hence, solve the equation:
log2(2x + 1) – 5 log4 x2 + 4 log2 x = 4

Solution:
(a)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x
= log 2 ( 2 x + 1 ) 5 log 2 x 2 log 2 4 + 4 log 2 x = log 2 ( 2 x + 1 ) 5 2 log 2 x 2 + log 2 x 4 = log 2 ( 2 x + 1 ) log 2 ( x 2 ) ( 5 2 ) + log 2 x 4
= log 2 ( 2 x + 1 ) log 2 x 5 + log 2 x 4 = log 2 ( 2 x + 1 ) ( x 4 ) x 5 = log 2 2 x + 1 x

(b)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x = 4
log 2 2 x + 1 x = 4    2 x + 1 x = 2 4    2 x + 1 x = 16    2 x + 1 = 16 x  14 x = 1 x = 1 14


Long Questions (Question 1 – 2)


5.6.1 Indices and Logarithms, SPM Practice (Long Questions)

Question 1:
(a)  Find the value of
i.   2 log2 12 + 3 log25 – log2 15 – log2 150.
ii.   log832
(b)  Shows that 5n  + 5n + 1 + 5n + 2  can be divided by 31 for all the values of n which are positive integer.

Solution:
(a)(i)
2 log2 12 + 3 log2 5 – log215 – log2 150
= log2 122 + log2 53– log2 15 – log2 150
= log 2 12 2 × 5 3 15 × 150
= log2 8
= log2 23
= 3

(a)(ii)
log 8 32 = log 2 32 log 2 8     = log 2 2 5 log 2 2 3 = 5 3

(b)
5n  + 5n + 1 + 5n + 2
= 5n  + (5 × 5n ) + (52 × 5n )
= 5n  (1 + 5 + 52)
= 31 × 5n  

Therefore, 5n  + 5n + 1 + 5n + 2 can be divided by 31 for all the values of n which are positive integer.



Question 2:
(a)  Given log10 x = 3 and log10y = –2. Shows that 2xy – 10000y2 = 19.
(b)  Solve the equation log3 x = log9(x + 6).

Solution:
(a)
log10x = 3   → (x = 103)
log10y = –2 → (y = 10-2)
2xy – 10000y2 = 19
Left hand side:
2xy – 10000y2
= 2 × 103 × 10-2 – 10000 (10-2)2
= 20 – 10000 (10-4)
= 20 – 1
= 19
= right hand side

(b)
log 3 x = log 9 ( x + 6 ) log 3 x = log 3 ( x + 6 ) log 3 9 log 3 x = log 3 ( x + 6 ) log 3 3 2 log 3 x = log 3 ( x + 6 ) 2
2log3 x= log3 (x + 6)
log3 x2= log3 (x + 6)
x2= x + 6
x2x – 6 = 0
(x + 2) (x – 3) = 0
x = – 2 atau 3.
log3 (– 2) not accepted (logarithm of a negative number is undefined)
Jadi, x = 3.

Short Questions (Question 12 – 14)


Question 12
Solve the equation,  log 2 5 x + log 4 16 x = 6

Solution:
log 2 5 x + log 4 16 x = 6 log 2 5 x + log 2 16 x log 2 4 = 6 log 2 5 x + log 2 16 x 2 = 6 2 log 2 5 x + log 2 16 x = 12 log 2 ( 5 x ) 2 + log 2 16 x = 12 log 2 ( 25 x ) + log 2 16 x = 12 log 2 ( 25 x ) ( 16 x ) = 12 log 2 400 x 2 = 12 400 x 2 = 2 12 x 2 = 10.24 x = 3.2




Question 13
Given that 2 log2 (xy) = 3 + log2x + log2 y
Prove that x2 + y2– 10xy = 0.

Solution:
2 log2 (xy) = 3 + log2x + log2 y
log2 (xy)2 = log2 8 + log2 x + log2y
log2 (xy)2 = log2 8xy
(xy)2 = 8xy
x2– 2xy + y2 = 8xy
x2 + y2 – 10xy = 0 (proven)



Question 14 (2 marks):
Given 2p + 2p = 2k. Express p in terms of k.

Solution:
2 p + 2 p = 2 k 2( 2 p )= 2 k 2 p = 2 k 2 1 2 p = 2 k1 p=k1

Short Questions (Question 9 – 11)


Question 9
Solve the equation,  log 2 4 x = 1 log 4 x

Solution:
log 2 4 x = 1 log 4 x log 2 4 x = 1 log 2 x log 2 4 log 2 4 x = 1 log 2 x 2 2 log 2 4 x = 2 log 2 x log 2 16 x 2 = log 2 4 log 2 x log 2 16 x 2 = log 2 4 x 16 x 2 = 4 x x 3 = 4 16 = 1 4 x = ( 1 4 ) 1 3 = 0.62996




Question 10
Solve the equation,  log 4 x = 25 log x 4

Solution:
log 4 x = 25 log x 4 1 log x 4 = 25 log x 4 1 25 = ( log x 4 ) 2 log x 4 = ± 1 5 log x 4 = 1 5    or    log x 4 = 1 5 4 = x 1 5     4 = x 1 5 x = 4 5 4 = 1 x 1 5 x = 1024   x 1 5 = 1 4     x = 1 1024



Question 11
Solve the equation,  2 log x 5 + log 5 x = lg 1000

Solution:
2 log x 5 + log 5 x = lg 1000 2. 1 log 5 x + log 5 x = 3 × ( log 5 x ) 2 + ( log 5 x ) 2 = 3 log 5 x ( log 5 x ) 2 3 log 5 x + 2 = 0 ( log 5 x 2 ) ( log 5 x 1 ) = 0 log 5 x = 2   or   log 5 x = 1 x = 5 2   x = 5 x = 25

Short Questions (Question 5 – 8)


Question 5
Solve the equation, log 9 ( x 2 ) = log 3 2

Solution:
log 9 ( x 2 ) = log 3 2 log a b = log c b log c a log 3 ( x 2 ) log 3 9 = log 3 2 log 3 ( x 2 ) 2 = log 3 2 log 3 ( x 2 ) = 2 log 3 2 log 3 ( x 2 ) = log 3 2 2 x 2 = 4 x = 6




Question 6
Solve the equation, log 9 ( 2 x + 12 ) = log 3 ( x + 2 )

Solution:
log 9 ( 2 x + 12 ) = log 3 ( x + 2 ) log 3 ( 2 x + 12 ) log 3 9 = log 3 ( x + 2 ) log 3 ( 2 x + 12 ) = 2 log 3 ( x + 2 ) log 3 ( 2 x + 12 ) = log 3 ( x + 2 ) 2 2 x + 12 = x 2 + 4 x + 4 x 2 + 2 x 8 = 0 ( x + 4 ) ( x 2 ) = 0 x = 4  (not accepted) x = 2




Question 7
Solve the equation, log 4 x = 3 2 log 2 3

Solution:
log 4 x = 3 2 log 2 3 log 2 x log 2 4 = 3 2 log 2 3 log 2 x 2 = 3 2 log 2 3 log 2 x = 2 × 3 2 log 2 3 log 2 x = 3 log 2 3 log 2 x = log 2 3 3 x = 27




Question 8
Solve the equation, 2 log 5 2 = log 2 ( 2 x )

Solution:
2 log 5 2 = log 2 ( 2 x ) 2 = log 5 2. log 2 ( 2 x ) 2 = 1 log 2 5 . log 2 ( 2 x ) 2 log 2 5 = log 2 ( 2 x ) log 2 5 2 = log 2 ( 2 x ) 25 = 2 x x = 23

Short Questions (Question 1 – 4)


Question 1
Solve the equation, log3 [log2(2x – 1)] = 2

Solution:
log3 [log2 (2x – 1)] = 2 ← (if log a N = x, N = ax)
log2 (2x – 1) = 32
log2 (2x – 1) = 9
2x – 1 = 29
x = 256.5




Question 2
Solve the equation,   l o g 16 [ l o g 2 ( 5 x   4 ) ] = l o g 9 3

Solution:
l o g 16 [ l o g 2 ( 5 x   4 ) ] = l o g 9 3 l o g 16 [ l o g 2 ( 5 x   4 ) ] = 1 4 log 9 3 = log 9 3 1 2 = 1 2 log 9 3 = 1 2 ( 1 log 3 9 ) = 1 2 ( 1 2 ) = 1 4 l o g 2 ( 5 x   4 ) = 16 1 4 l o g 2 ( 5 x   4 ) = 2 5 x   4 = 2 2 5 x = 8 x = 8 5



Question 3
Solve the equation, 5 log 4 x = 125

Solution:
5 log 4 x = 125 log 5 5 log 4 x = log 5 125 put log for both side ( log 4 x ) ( log 5 5 ) = 3 ( log 4 x ) ( 1 ) = 3 x = 4 3 = 64




Question 4
Solve the equation, 5 log 5 ( x + 1 ) = 9

Solution:
5 log 5 ( x + 1 ) = 9 log 5 5 log 5 ( x + 1 ) = log 5 9 log 5 ( x + 1 ) . log 5 5 = log 5 9 log 5 ( x + 1 ) = log 5 9 x + 1 = 9 x = 8