# Mathematics Grade 12 Paper 2 Eastern Cape June 2014 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Eastern Cape June 2014, 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-p2-june-2014-grade-12-memo/a8d1d2e3-fa48-4cb0-b659-3384a72d37b2
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2014
- Exam period: Eastern Cape June
- Paper: 2
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 12 · Eastern Cape June · 2014. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

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Mathematics/P2                           1                      DBE/November 2011
                                  NSC – Memorandum

                              MATHEMATICS P2

                         JUNE 2014 – COMMON TEST

                                MEMORANDUM

                               NATIONAL
                          SENIOR CERTIFICATE

                                   GRADE 12

MARKS: 125

                        This memorandum consists of 11 pages.

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Mathematics/P2                                      2                  Common Test June 2014
                                             NSC – Memorandum

QUESTION 1

1.1.1      AC2 = (–1 – 4)2 + (–4 – 1)2 = 9 + 9 = 45                     correct
                                                                       substitution into
           AC =      18                                                distance formula
                                                                        18
                 = 3 2                                          (3)    answer

1.1.2           1  2 4  2    1 
           M is       ;        ; 1                         (2)    answer
                 2      2   2 
1.1.3                        2 4 6                                   mAB  2
           Gradient of AB =             2
                               2 1  3
                                                         1             m bisectorof AB  
                                                                                                1
                            gradient of  bisector is                                          2
                                                         2
           (obviously it passes through M)
                       1 
                       ; 1
                       2 
                              1
                       y xc
                              2
                            1  1
           Substitute 1       c                                   substitution of m
                            2  2                                   and M into
                                1                                      equation of line
                      c 1
                                4
                             3
                          
                             4

                            1    3
           Equation is y =  x                                 (4)
                            2    4                                     answer

1.2        x 2  2 x  1  y 2  2 y  1 2 x  2 y
                                                                        x2 – 4x + 4
           x2  4x  y 2  4 y   2                                    y2 + 4y + 4
           x2  4x  4  y 2  4 y  4   2  4  4                    (x – 2)2
                                                                        (y + 2)2
           x  22   y  22  6
                                                                        answer: centre
            centre is (2; –2) and radius =       6             (6)     answer: radius

                                                                [15]

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Mathematics/P2                                              3                               Common Test June 2014
                                                     NSC – Memorandum

QUESTION 2
2.1.1      Since JK // LM, MJK = MLM                                                     equating gradients
                4 1          02
                                                                                        substitution in
                p2            53
                                                                                         each gradient
                  2  p  2    5 2
                  2 p  4   10                                                  (4)   simplification
                  2p  6
                 p3
2.1.2        JK2  3  2   4  1  50                                               correct substitution
                                2                    2

                                                                                           into distance form.
             LM 2  2  0  3  5  8
                                    2            2

                                                                                   (5)    50
             JK : LM          50 : 8  5 2 : 2 2                                         8
                                             5:2                                         ratio
                                                                                         answer
2.1.3      Diagonals of a parallelogram bisect each other
                                                                                            3 1
                            3 1                                                         ; 
           Midpoint of JL :  ;                                                            2 2
                            2 2
                                                                                            x 3     3
           This is the same midpoint for MQ:                                                      
                                                                                              2      2
                  x 3     3             y  2     1                                        y  2       1
           Thus                ;                                                                
                    2      2               2       2                                          2         2
           x  0                y  1                                                 answer OR answer
           Therefore Q is Q(0; –1)                                                (5)    only(if done by
                                                                                         inspection) – full
                                                                                         marks
2.1.4      Since the x coordinates of K and M are both 3, it follows the
           equation of KM is x = 3.                                                      answer
                                                                      (2)
              o
2.1.5       90 since KM is a vertical line                            (1)                 answer

2.1.6      For collinearity;            mJR  mJL                                         equating: mJR  mJL
                                                       k 1              0 1                        k 1
                                        mJR =                       =                     mJR =
                                                     1  ( 2)          5  (2)                    1  ( 2)

                                                     k 1               1               simplification
                                             =                      =                    answer
                                                       3                7

                                                               3
                                         k – 1 =
                                                               7
                                                         3
                                         k =               + 1
                                                         7

                                                               4
                                                         k =                       (4)
                                                               7

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Mathematics/P2                               4                     Common Test June 2014
                                      NSC – Memorandum

2.2.1      Q (x; 2) … (radius QR  tangent)                      yQ  2
           Substitute (x; 2) in 3x + 4y + 7 = 0:
                               3x + 8 + 7 = 0                   substitution: y = 2
                           x = –5                                x = –5
                         Q (–5; 2)
           Radius = QR = 0 – (–5) = 5                            radius
            Equation is (x + 5)2 + (y – 2)2 = 25        (5)     equation
2.2.2      QR   =       5 units
           d    =       2 x radius
            WZ =       10 units                         (1)
                                                         [27]
                                                                answer

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Mathematics/P2                                                5                   Common Test June 2014
                                                       NSC – Memorandum

QUESTION 3

3.1      sin 15o = sin (45o – 30o)                                                15° as 45 – 30
                 = sin 45o cos 30o – cos 45o sin 30o                               correct
                    1     3    1 1                                                expansion
                 =     .         .
                     2 2        2 2
                                                                                   correct special
                         3      1
                     =                                                             angle values
                       2 2 2 2
                        3 1      2
                     =        x
                        2 2       2
                             6 2
                     =                                                    (3)
                              4

                                                       OR

         sin 15o = cos 75o
                 = cos (45o + 30o)
                =cos 45o cos 30o – sin 45o cos 30o                                75° as 45 + 30
                       1   3   1   1                                     correct
                                      
                       2   2   2   2 
                                                                                  expansion
                =
                                                                                   correct special
                                    3   1                                           angle values
                     =                
                                  2 2 2 2
                                  3 1         2
                     =                 x
                                  2 2          2
                                  6 2
                     =                                                    (3)
                                   4

3.2                                           
                       tan 180 o   . cos 360 o              
                                                        
         sin 180 o   . cos 90 o    cos 540 o   . cos                  For each reduction
                                                                                  :
                            tan  . cos  
                                                                                   tan   cos 
                sin   .  sin    cos  . cos                               sin  – sin 
                   sin                                                           – cos cos 
                            cos 
                   cos                                                                     sin 
         =                                                                         tan  
                 sin 2   cos 2                                                          cos 
                        sin 
         =
                         
                 sin   cos 2 
                       2
                                                                                  sin 2   cos 2 
         =     – sin θ                                                      (9)
                                                                                  answer
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Mathematics/P2                                          6                                  Common Test June 2014
                                                 NSC – Memorandum

3.3
                               
       sin 45o  x . sin 45o  x       12 cos 2 x
       LHS                  sin 45  x  . sin 45  x 
                                         o           o

                            sin 45 cos x  cos 45 sin x  sin 45 cos x  sin x cos 45 )  expansion
                                         o               o                            

                              1             1             1         1        
                                . cos x       . sin x  .  cos x     sin x               special ∠
                              2               2           2          2                      values
                             1            1
                              cos 2 x  sin 2 x
                             2            2                                                     difference of

                                                
                             1                                                                   squares
                               cos 2 x  sin 2 x                                                         1
                             2                                                                  HCF of
                             1                                                                            2
                              cos 2 x
                             2
                            RHS                                                     (5)

3.4              sin 33o cos 33o
                        
                 sin 11o cos 11o
                 sin 33o. cos 11o  cos 33o .sin 11o                     sin 33o. cos 11o  cos 33o . sin 11o
      =
                           sin 11o .cos 11o                              sin 11o.cos 11o
      =
                     
                 sin 33o  11o                                               
                                                                         sin 33o  11o    
                 sin 11o cos 11o
                     sin 22o                                             sin 22o
      =
                 sin 11o cos 11o
                 2 sin 11o cos 11o                                       2 sin 11o cos 11o
      =                      o               o
                   sin 11 cos 11

      =        2                                                (6)      answer

3.5     tan 3x
                1
       tan 24 o                                                          tan 24°

       tan 3x  tan 24 o                                                 tan 3x  tan 24o
      3x  24 o  k . 180 o                                              3x  24o  k . 180o

       x  8 o  k . 60  , k                                (5)   x  8  k . 60 , k  
                                                                             o        

                                                               [28]

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Mathematics/P2                                                7                        Common Test June 2014
                                                       NSC – Memorandum

QUESTION 4

4.1                    F                     y

                              h                    g

                       Q                     G         f        H         x

        Constr: Draw FQ  HG produced                                            construction

                 sin FĜH = sin(180° – FĜQ ) = sin FĜQ                         reduction

                 FQ
                                                sin G                              FQ
                  h                                                                    sin G
                  FQ                =           h sin G                             h
                 FQ                                                                 FQ
        Also                                    sin H                                 sin H
                  g                                                                  g
                  FQ                =           g sin H
                  h sin G           =           g sin H                         h sin G = g sin H
                   s in G                         s in H
                                        
                      g                              h
                                                                          (5)
4.2.1            PW
                                        tan α 
                 PQ                                                              using tan α
                  PW                   PQ tan α                              answer
                                                                          (2)
4.2.2   QP̂R  180o  x  y  and                                               RP̂Q  180o  x  y 
                              
        sin QP̂R  sin 180o   x  y           
                        sin  x  y 
        In Δ PQR, by the sine rule
          PR       QR
                                                                                choice of sine formula
         sin Q̂ sin QP̂R
          PR           15                                                        correct substitution into
              
                        
         sin y sin 180  x  y 
                      o
                                                                                 sine formula
                      15 sin y                                                   sin x  y 
         PR 
                     sin  x  y 
                                                                          (4)

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Mathematics/P2                                      8                          Common Test June 2014
                                             NSC – Memorandum

4.2.3     PW         =    PR tan α from 3.3.1
                                                                                  15 sin y
                 
                           15 sin y
                                        . tan                          PW                  . tan 
        PW
                          sin  x  y                                           sin x  y 

                            15 sin y
                                       . tan 
                          sin  y  y 

                          15 sin y
                                  . tan 
                           sin 2 y                                      simplification

                             15 sin y
                                        . tan 
                          2 sin y. cos y                                sin 2y = 2 sin y. cos y

                            7 ,5
                                 . tan 
                          cos y
                                 tan α
         PW         =    7,5
                                 cos y
                                                                 (3)
                                                                [14]

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Mathematics/P2                                     9                          Common Test June 2014
                                            NSC – Memorandum

QUESTION 5

5.1      Supplementary                                                (1)    Answer

5.2.1    It is the angle between tangent and radius.                  (1)    Answer

5.2.2    Ŝ3  Ŝ4  90o ... Tan  radius
         N̂1  N̂2  90o … Tan  radius                                     Statement Reason
         Ŝ3  Ŝ4  N̂1  N̂2  90o  90o                                   Statement
                               = 180°
          RNOS is a cyclic quadrilateral … opp  quad supplementary         Reason
                                                                (4)
5.2.3    Ŝ1               x
         S1                N̂2  x … Tan chord theorem                     Statement Reason
         N̂2                S3  x … base ' s of isosceles ∆ OSN          Statement Reason
         Ŝ3               R̂ 2  x … ' s in same segment                 Statement Reason
         N̂2                          ’s in same segment. 
                             R̂  x...
                                1                                             Statement Reason
                                                                        (8)

5.2.4    Ô1  Q̂2  N̂2  Ŝ3  180o … sum of ' s of Δ OSN                 Statement with
                                                                              reason
         But Ŝ3  N̂ 2 … s opp. equal sides                                Statement with
          Ô1  Ô2  2Ŝ3  180o                                            reason
         Ô1  Ô2  O3  180o … ' s on a straight line                     Statement with
                                                                              reason
          Ô1  Ô2  2 Ŝ3  Ô1  Ô2  Ô3
         2 Ŝ3  Ô3                                                         2 Ŝ3  Ô3
                1
          Ŝ3  Ô3                                                   (4)
                2

                                             OR

                                 ext. s of cyclic quad. 
                     Ô3  SR̂N ....                                          Statement Reason

                 R̂ 2  Ŝ3 … ∠s in the same segment                         Statement with
         but     SO = ON … radii of a circle                                  reason
                R̂1  R̂ 2 ...  chords ;  s                              Statement with
                            1                                                 reason
                    Ŝ3  Ô3
                            2
                                                                        (4)
                                                                       [18]

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Mathematics/P2                                10                               Common Test June 2014
                                       NSC – Memorandum

QUESTION 6

6.1        Given: ∆ABC with D
                                                    A
           on AB and E on
           AC such that DE//BC
                                                  h1 h2

                                           D               E

                                   B                              C
                     AD   AE
           RTP:         
                     DB   EC

           Construction: Join DC and BE                                         construction

           Proof:
                        1                areas of ∆’s with the same
                          AD.h1                                                  Statement Reason
           area Δ ADE   2          AD
                                    … height and common vertex                NOTE: If area of 
            area BDE    1          DB    are in the same ratio as their
                          DB.h 2                                                 found, then reason
                        2                bases                                  not necessary
                        1
                          AE.h 2
           area Δ ADE   2          AE
                                                                              Statement
            area DEC    1          EC
                          EC.h 2
                        2

           but area ∆ DBE = area ∆ DEC ... Same base, same parallel lines      Statement Reason
                 AD AE
           Thus                                                         (6)
                 DB EC

6.2.1      In ΔPQM, GH ∥ QT
           QH   GP                              
              =     (line parallel to one side of a Δ OR Prop Th: GH ∥ PQ)      Statement Reason
           HM GM
                  1                                                              Answer
              =                                                       (3)
                  2
6.2.2      QH = k ; HM = 2k  RM = 3k                                           RM = 3k
           MR = QM = 3k ...M is the midpoint of QR
           RG RH                                                  
              =      (line parallel to one side of a Δ OR Prop Th: GH ∥ PQ))     Statement Reason
           RT RQ
                 5k                                                              Substitution
              =                
                 6k
                 5
              =                                                         (5)     answer
                 6

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Mathematics/P2                                   11                          Common Test June 2014
                                          NSC – Memorandum

6.3.1      Let Ẑ2  x  α …. Tan chord theorem                              Statement Reason

           Then AB̂X = 90o – α … sum of ' s of ∆ ABP                         Statement with
                                                                               reason
           But Ẑ1  AB̂P
                   = 90 – α ... ' s opposite equal sides:AZ = AB             Statement with
                                                                               reason
           Ẑ1  Z2  α  90o  α … adj. ' s on a straight line              Statement with
                    = 90o                                                      reason
           Thus Ẑ 3 = 90o
                                                                      (5)
6.3.2                In ∆ AYZ and ∆ AZX
           1.         Ẑ2  X̂ … Tan chord theorem                            Statement with
                                                                               reason
           2.        Â2          Â2     ….common                           Statement
           3.        AŶZ         AẐX (remaining angles)
                                                                               Statement with
                      Δ AYZ /// Δ AZX         , ,                (3)      reason

6.3.3        AZ             AX                                                 statement
                                s ∥∣ sides in proportion 
             AY             AZ
            AZ2           AY . AX                                    (1)
                                                                      [23]

                                 TOTAL MARKS: [125]

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- [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr
- [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng
- [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.