Practice Quiz : Maths 38|Class :12th | pages:07

Mathematics/P2 6 DBE/November 2010

Questions and Answers
NSC

QUESTION 5 1
In the diagram below, A, B and C are the vertices of a triangle. AC is extended to cut the x-axis
at D.

y
A(3 ; 7)

C(– 3 ; 4)

G(a ; b)

D x
0
M ●

B(1 ; –6)

5.1 Calculate the gradient of:

5.1.1 AD (2)

5.1.2 BC (1)

5.2 Calculate the size of DĈB . (3)

5.3 Write down an equation of the straight line AD. (2)

5.4 Determine the coordinates of M, the midpoint of BC. (2)

5.5 If G(a ; b) is a point such that A, G and M lie on the same straight line, show that
b = 2a + 1. (4)

5.6 Hence calculate TWO possible values of b if GC = 17 . (6)

[20]

Copyright reserved Please turn over

ANSWERS
Mathematics/P2 22
2NSC – Memorandum
7 PracticeQuiz
Practice Quiz : Maths
: Maths
DBE/November 2010 3838
QUESTION 5
y
A(3 ; 7)

C(−3 ; 4)

G(a ; b)

D β P α x
(x ; 0) O
M

B(1 ; −6)
5.1.1 mAD = m AC mAD = m AC substitution of A and
Note: If candidate C into correct formula
7−4 4−7
= = gives
3 − (−3) − 3 − (3) 7
3 OR −3 mAD =
= = 3− x
6 −6 answer
1 1 then 1/2 marks (2)
= =
2 2

5.1.2 −6−4 4 − ( − 6)
m BC = m BC =
1 − (−3) − 3 − (1)
− 10 10
= OR =
4 −4
−5 −5 answer
= =
2 2 (1)
5.2 1
m AD = = tan CD̂O
2
26,57°
CD̂O = 26,56505....°
−5
m BC = = tan α
2
α = 111,814 095° 111,80°
DĈB = 111,801 409 5....° − 26,56505....°
= 85,236359°
= 85,24°
≈ 85,2° answer
(3)

OR

Copyright reserved Please turn over

 3
Mathematics/P2 8 DBE/November 2010
NSC – Memorandum
1
tan CD̂O =
2
CD̂O = 26,56505....°
26,57°
5
tan(180° − α ) =
2
180° − α = 68,19859051...° 68,2°
DĈB = 180° − (26,56505....° + 68,19859051...°)
= 85,236359°
= 85,24° answer
OR (3)

DĈB = α − CD̂O
m − mCD
tan DĈB = CB
1 + mCB .mCD mCB − mCD
tan DĈB =
− 52 − 12 1 + mCB .mCD
=
1 + (− 52 )( 12 ) substitution
= 12
answer
DĈB = 85,24° (3)

OR
AC = 45 BC = 116 AB = 173
AC 2 + BC 2 − AB 2
cos AĈB =
2AC.BC cosine rule
45 + 116 − 173
= substitution into
2( 45 )( 116 )
cosine rule
= −0,083045...
AĈB = 94,76...°
DĈB = 180° − 94,76...°
= 85,24° answer
(3)
OR
D(– 11 ; 0)
DC = 80 BC = 116 DB = 180
DC 2 + BC 2 − DB 2 cosine formula
cos DĈB =
2DC.BC
80 + 116 − 180 substitution into
= cosine rule
2( 80 )( 116 )
= 0,08304547985...
DĈB = 85,24° answer
(3)

OR

Copyright reserved Please turn over

 4
Mathematics/P2 9 DBE/November 2010
NSC – Memorandum

Equation AC: 2 y = x + 11
D(– 11 ; 0)
C(– 3 ; 4)
DC 2 = ( xC − x D ) 2 + ( y C − y D ) 2
= (−3 + 11) 2 + (4 − 0) 2
= 80
Equation BC: 2 y = −5 x − 7
P (− 75 ;0)
PC 2 = (−3 + 75 ) 2 + (4 − 0) 2
464
=
25
DP = (− 75 + 11) 2
2

2304
=
25
In ΔDCP: DP 2 = DC 2 + CP 2 − 2DC.CP. cos DĈP cosine formula
2304 2000 464 ⎛ 2000 ⎞⎛ 464 ⎞
= + − 2⎜⎜ ⎟⎜ ⎟
⎟⎜ 5 ⎟. cos DĈP substitution into
25 25 25 ⎝ 5 ⎠⎝ ⎠ cosine rule
DĈP = 85,23635...
DĈP = 85,24° answer
(3)

Copyright reserved Please turn over

 5
Mathematics/P2 10 DBE/November 2010
NSC – Memorandum

5.3 1 substitution of
y−7 = (x − 3) (3 ; 7) into
2
1 11 y − y1 = m(x − x1 )
y = x+
2 2
answer in any
x − 2 y + 11 = 0
form
(2)
OR
1
y−4= (x + 3) substitution of
2 (– 3 ; 4) into
1 11 y − y1 = m(x − x1 )
y = x+
2 2
x − 2 y + 11 = 0 Note: answer in any
If candidate leaves answer as form
1 1
OR y − 7 = (x − 3) or y − 4 = (x + 3) : (2)
1 2 2
y= x+c max 1 / 3 marks
2
substitution of
1 (3 ; 7) into
(7) = (3) + c
2 y = mx + c
11
c=
2
1 11
y = x+ answer in any
2 2
form
x − 2 y + 11 = 0 (2)

5.4 ⎛ − 3 +1 4 − 6 ⎞ substitution
M ( x; y ) = ⎜ ; ⎟
⎝ 2 2 ⎠
answer (2)
M ( x; y ) = (− 1; − 1)

Copyright reserved Please turn over

 6
Mathematics/P2 11 DBE/November 2010
NSC – Memorandum
5.5 7 − (−1) gradient = 2
m AM = =2
3 − (−1)
substitution
y = 2x + c Note: (– 1 ; – 1)
− 1 = 2(−1) + c If the candidate does not conclude
c=1
b = 2a + 1 from y = 2x+ 1:
∴c = 1
max 3 / 4 marks conclusion
y = 2x + 1
(4)
G(a ; b) lies on the line
∴ b = 2a + 1

OR 7−b
7 − b b +1
= 3− a
3 − a a +1 b +1
(7 − b)(a + 1) = (b + 1)(3 − a )
a +1
7a + 7 − ab − b = 3b − ab + 3 − a
8a − 4b = −4 equating
2a − b = −1
simplification
b = 2a + 1 leading to
2 a − b = −1
OR (4)
Using the point (– 1 ; – 1) substitution of
b +1 8 (– 1 ; – 1) into
=
a +1 4 gradient
b +1 gradient = 2
=2
a +1 equating
b + 1 = 2a + 2 simplification
leading to
b = 2a + 1 b + 1 = 2a + 2
(4)
OR

Using the point (3 ; 7) substitution of

7−b 8 (3 ; 7) into
=
3− a 4 gradient
7−b gradient = 2
=2
3− a equating
7 − b = 6 − 2a simplification
leading to
b = 2a + 1 7 − b = 6 − 2a
(4)

Copyright reserved Please turn over

 7
Mathematics/P2 12 DBE/November 2010
NSC – Memorandum
5.6 GC = 17
GC 2 = 17
(a + 3) 2 + (b − 4) 2 = 17 distance
Note: formula in terms
(a + 3) 2 + (2a + 1 − 4) 2 = 17 If candidate swops a of a and b
a 2 + 6a + 9 + 4a 2 − 12a + 9 − 17 = 0 and b around: substitution of
5a 2 − 6 a + 1 = 0
max 2 / 6 marks b = 2a + 1
standard form
(5a − 1)(a − 1) = 0 factors or
1 correct
a= or a = 1
5 substitution into
7 formula
∴b = or b = 3 values of a
5 values of b
(6)
OR

b −1
a= b −1
2 a=
2
17 = (a + 3) 2 + (b − 4) 2 distance
⎛⎛ b −1⎞ ⎞
2 formula in terms
17 = ⎜⎜ ⎜ ⎟ + 3 ⎟⎟ + (b − 4)
2
of a and b
⎝⎝ 2 ⎠ ⎠ substitution
⎛b +5⎞
2
b −1
17 = ⎜ ⎟ + (b − 4)
2 of a =
⎝ 2 ⎠ 2
b + 10b + 25 + 4b 2 − 32b + 64
2
17 =
4
68 = 5b − 22b + 89
2
standard form
factors or
0 = 5b 2 − 22b + 21
correct
0 = (5b − 7)(b − 3) substitution into
7 formula
∴b = or b = 3 values of b
5
(6)
[20]

Copyright reserved Please turn over