Pdf Level 3 Unit 2 P2

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NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 FEBRUARY/MARCH 2009 MEMORANDUM This memorandum consists of 13 pages. Copyright reserved Please turn over Mathematics/P2 2 NCS – Memorandum DoE/Feb. – March 2009 QUESTION 1 1.1 m m 1 − 0 6 − 3 1 = 3 = BC substitution into gradient formula answer (2) BC 1.2 m AD = m BC m AD = 1 3 m AC = ----------------------------- AB//BC 1 3 ∴ Equation of AD is: 1 y= x+c 3 1 6 = (1) + c 3 17 c= 3 1 17 ∴y = x+ 3 3 OR 1 y − 6 = ( x − 1) 3 1 1 y−6= x− 3 3 1 17 y= x+ 3 3 1.3 1 17 x+ 3 3 1 17 t = (7 ) + 3 3 t =8 y= substitution of (1 ; 6) into a straight line equation equation (3) m AC = 1 3 substitution of (1 ; 6) into a straight line equation equation (3)…show more content…
(2) OR t −6 1 = 7 −1 3 t −6 = 2 ∴t = 8 Copyright reserved Please turn over Mathematics/P2 3 NCS – Memorandum DoE/Feb. – March 2009 1.4 AD = (8 − 6) 2 + (−1 − 3) 2 AD = 40 AD = 2 10 BC = (6 − 3) 2 + (1 − 0) 2 BC = 10 AB = (6 − 0) 2 + (1 − 3) 2 AB = 40 AB = 2 10 using distance formula answer for AD answer for BC answer for AB (4) 1.5 m AB = m AB 6−0 1− 3 = −3 = m AB = −3 1 − 0 1 = 6 − 3 3 1 m AB .m BC = × −3 3 = −1 ∴ AB ⊥ BC m BC m AB × mBC = −1 conclusion (3) formula for area of ∆ 1 1 2 10 2 10 + 2 2

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