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LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 1
MODULE 13(A) ADDITIONAL MATHEMATICS
TOPIC : LINEAR LAW 1.
Diagram below shows a straight line graph of 2
y
x against x.
Given that 2 32y x x= + , calculate the values of h and k.
2.
The diagram above show the graph of 2log y against x. The variables x and y
are related by the equation x
ay
b= where a and b are constants. Find the
values of a and b.
3.
Diagram shows the graph of straight line obtained by plotting 2
y
x against x.
Find (a) y in terms of x, (b) value of y when x=1
2
y
x
x
(h,3)
(6,k)
2log y
x
3
(4,-5)
2
y
x
x
3
-5
LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 2
4.
Diagram show a linear graph of 1
y against 2x . The variables x and y are
related by the equation 22p
x qy
= + , where p and q are constants.
(a) determine the values of p and q, (b) express y in terms of x.
5.
x and y are related by the equation q
y pxx
= + , where p and q are constants. A
part of line of best fit is obtained by plotting y
x against
2
1
x as shown in
diagram.
Calculate the values of p and q.
6.
Diagram shows the graph of a straight line 3log y against 3log x . Express y in
terms of x.
1
y
x²
-2
(4,6)
y
x
2
1
x
(2,5)
(4,9)
3log y
3log x
-2
(3,10)
LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 3
7.
The variables x and y are related by the equation 2axy b x+ = , where a and b
are constants. A straight line is obtained by plotting 2y against 1
x. Find the
values of a and b.
8. Given x and y are related by the equation
qxy px
x= + where p and q are
constants. A straight line graph is obtained by plotting y against 2
1
x, it passes
through the points (0 , 0.5) and (4 , -2.5). Calculate the value of p and q.
9.
The variables x and y are related by the equation 2y kx= where k is a
constant.
(a) Convert the equation 2y kx= to linear form.
(b) Find the value of (i) lgk
(ii) h.
10.
Figure shows a straight line graph of y
x against x. Given that 25 2y x x= − ,
calculate the value of p and of q.
END OF MODULE
lgy
lg x
6
(h,2)
y
x
x
(q,2)
(3,p)
2y
1
x
(3,5)
(1,2)
LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 4
ANSWERS MODULE 13(A)
ADDITIONAL MATHEMATICS TOPIC : LINEAR LAW
1.
22 , 1
31 3 2
6
1 , 8
yx m
x
kor h
h
h k
= + =
−= = +
−
= =
2.
2 2 2
2 2
log log log ( )
3 , 2
log 3 , log 2
8 4
y a b x
C m
a b
a b
= −
= = −
= =
= =
3. (a)
2
3 2
3
5
33
5
33
5
m
yx
x
y x x
=
= +
= +
(b) 3 23
(1) 3(1)5
33
5
y = +
=
4. (a)
2
2 , 2
1 2
22
1
2
2
m c
qx
y p p
p
p
q
p
q
= = −
= +
=
=
= −
= −
(b)
2
1
2 2y
x=
−
LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 5
5.
2
2
2
5 2(2)
1
m
y qp
x x
q
p
p
=
= +
=
= +
=
6.
3 3
23 3 3
4
4 , 2
log 4log 2
log 4log log 3
9
m c
y x
y x
xy
= = −
= −
= −
=
7.
( )
2 1 1
3
2
(1,2)
3 12 1
2
2
3
by
a x a
bm
a
a
a
b
= − +
= − =
= +
=
= −
8.
( )
2
3,
4
3
4
30.5 0
4
0.5
m c p
qy p
x
q m
p
p
= − =
= +
= = −
= − +
=
9. (a)
10 10 10log 2log logy x k= − +
(b) (i)
10log 6k =
(ii) 2
2 62
0
2
m
h
h
= −
−= −
−
=
LINEAR LAW FORM 5
PROGRAM KECEMERLANGAN AKADEMIK (szk smkpn) Page 6
10. 5 2
(3, ) ( ,2)
5 2(3) 2 5 2( )
31
2
yx
x
p q
p q
p q
= −
= − = −
= − =
END OF MODULE