Mathematics and Mathematical Literacy - Thutong

Mathematics(NCS)/Grade 12/ P2 93 ExemplarMEMORANDUMAB = 4 + 16Appropriate scale Correct drawingCorrect min and max values for the range Correct values for the median etc The values can be out by 1unit on eitherside. 52.1.1 Let the centre = (a; b)a = 2 22( x − 2) + ( y − b) 2 = rSub in (2;0)22 2( 2 − 2) + ( 0 − b ) = r 2 2∴ b = r Sub in (4; -6)22 2( 4 − 2) + ( − 6 − b ) = b 2 2∴4 + 36 + 12b + b = b ∴40 + 12b = 0∴ b = − 103 ⎛ 10 ⎞Centre = ⎜2; − ⎟⎝ 3 ⎠2.1.22 ⎛ 10 ⎞( x − 2) + ⎜ y + ⎟ = 4 ⎝ 3 ⎠1− 3 − ( − 6)m MB =3= - 4 2 − 4 3 ∴M tangent = 3 4 Tangent: y = 3 4 x + cSub in (4;-6)-6 = 3 4(4) + c-9 = c2Tangent: y = 43 x − 942.2.1 y = 5 − 2x2.2.2xx222( 5 − 2x) −12x− 6( 5 − 2 ) + 20 = 0+ x2+ 25 − 20x+ 4x−12x− 30 + 12x+ 20 = 05x2 − 20x + 15 = 0 x2− 4x + 3 = 0∴ ( x − 3)( x −1)= 0∴ x = 3 x = 1() 3y = 5 − 2 y = 5 − 2 1 ∴ y = −1 y = 3 ∴Points of intersection are:(3;-1) and (1:3) 6AB = ( 3 1) 2 + ( −1−3) 2()− 372.2.3AB = 20AB = 2 5 m BC = -1 - 33 -1 = -4 2 = -2 m perp = 1 2 ⎛1+3 3 −1⎞Midpoint of AB = ⎜ ; ⎟ = (2;1) ⎝ 2 2 ⎠Perpendicular bisector: y = 1 2 x + cSub (2;1): 1= 1 2(2) + c∴c = 0∴ y = 1 2 x 52.2.4 The x-intercepts of the circle are found by:2x −12x+ 20 = 0 ∴ ( x −10)( x − 2)= 0∴x = 10 or x = 2∴ x = 6 is the perpendicular bisector of thex-intercepts∴ the x value of the centre = 6∴ y = 3The centre of the circle = (6;3) 63.1.1 (5;1) p is 1 unit from C to the line x = 4,so the point C' will be 1 unit from x = 4 onthe other side i.e. 5. The y-value (q) remainsthe same. 33.1.2 (12;4) r is 3 units from C' to the line x= 9, so the point C″ will be 3 units from x =9 on the other side i.e. 12. The y-value (s)remains the same. 33.1.3 A translation 10 units right. TriangleABC has remained in the same horizontalplane but has moved 10 units along. 23.1.4 If point A (1;3) is reflected about the x = 9, itwill become A' = (17;3). If A’ is thenreflected about the x = 4 line, it will becomeA″ = (-9;3). This is not the same result asabove. 33.2.1 If A = (4;3) then A' = (3;4) and A″ =(-3;4) 33.2.2 Rotation of 90° :A″ = (4cos 90°− 3sin 90°; 4sin 90° + 3cos90°) A″ = (4 × 0 − 3 × 1; 4 × 1 + 3 × 0) A″ = (-3;4) 44.1.1 sin 2θcosθ+ cos 2θsin θ 21+2 2 cos θ −1( )2 sin θ cosθcosθ+ (2 cos24 cos θ −1sinθ(2 cos θ + 2 cos24 cos θ −1222θ −1) sinθ θ −1)6Copyright reserved