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National Senior Certificate

Mathematics

Paper 2 MEMORANDUM

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Grade 12

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–1 + 3 3 + (–1) M = _; _ ✓ 2 2 M = (1; 1) ✓ 4 + (–2) 4 + (–2) Midpoint FH = _; _ ✓ 2 2

2010 Mathematics Grade 12 Paper 2: Memorandum

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4 – (–2) 6 mFH = _ = _ = 1 ✓ 4 – (–2) 6 \ mEG × mFH = –1 Þ EG^FH ✓ \ using 1.2, diagonals of EFGH bisect at 90° ✓ \ EFGH is a rhombus. OR ____________________ lengthEH = √ (–1 – (–2))2 + (3 – (–2))2 ______ = √1 + 25 ___ = √26 ✓ ________________ lengthEF = √(–1 – 4)2 + (3 – 4)2 ______ = √25 + 1 ___ = √26 ✓ \ using 1.2, EFGH is a parallelogram (diagonals bisect) ✓ But EH = EF ✓ \ EFGH is a rhombus. mEG = –1 from above ✓ \ y = –x + c Substitute (–1; 3): ✓ 3 = –(–1) + c 2=c \ y = –x + 2 ✓ 5 y = –x + 2 Let x = _. 2 5 y = –_ + 2 ✓ 2 1 y = –_ ✓ 5 –3 \ _; _ does not lie on the line. ✓ 2 4 4 – (–1) 5 mFH = _ = _ = 5 ✓ 4–3 1 \ tan a = 5 A = tan–1(5) ✓ = 78,69° ✓

= (1; 1) ✓ \ midpoint FH = midpoint EG \ lines bisect each other. 3 – (–1) 4 mEG = _ = _ = –1 ✓ –1 – 3 –4

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2010 Mathematics Grade 12 Paper 2: Memorandum 1.7 lengthEG = √(3 – (–1))2 + (–1 – 3)2 ___ = √32 ✓ ___________________ lengthHM = √(1 – (–2))2 + (1 – (–2))2 ___ = √18 ✓ ___ ___ 1 \ area ▵EGH = _√ 18 × √ 32 ✓ 2 = 5,05 units2 ✓ G ® H back 5 down 1 \ E ® P is the same ✓ \ P(–6; 2) ✓ (or equivalent) ▵EDC is right angled at C (tangent, radius) ED2 = EC2 + DC2 ✓ 132 = 122 + DC2 25 = DC2 5 = DC ✓ DC2 = (a – 1)2 + (2 – (–1)2 ✓ 25 = a2 – 2a + 1 + 9 ✓ 0 = a2 – 2a – 15 ✓ a = –3; a = 5 ✓ By inspection, for this sketch a = 5. ✓ 2 – (–1) 3 m =_ = _ ✓

4 5–1 –4 ✓ _ mtangent = 3 _ y = –4 x + c Substitute (1; –1): 3 _ –1 = –4 x + c ✓ 3 1 _=c✓ 3 1 _ y = –4 x + _ 3 3

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