GRADE 12 MATHEMATICS LEARNER NOTES - Mail & Guardian

SENIOR SECONDARY ,03529(0(17 PROGRAMME 201 GRADE 12MATHEMATICSLEARNER NOTESThe SSIP is supported byc) Gauteng Department of Education, 20131

TABLE OF CONTENTSLEARNER NOTESSESSION TOPICRevision of Analytical Geometry (Grade 11)PAGE32c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)SESSION 15TOPIC: REVISION OF ANALYTICAL GEOMETRY (GRADE 11)Learner Note: Analytical Geometry is an important topic that carries a lot of marks in thematric final exam. Make sure that you know the basic formulae and then practise lots ofexamples involving applications of these formulae. The properties of quadrilaterals areextremely important in Analytical Geometry. Make sure you can prove that a quadrilateralis a parallelogram, rectangle, square, rhombus or trapezium by knowing the properties ofthese quadrilaterals.SECTION A: TYPICAL EXAM QUESTIONSQUESTION 1:15 minutesIn the diagram, PQRS is a trapezium with vertices P(5; 2), Q(1; 1), R(9; 5) and S , andPS//QR. PT is the perpendicular height of PQRS and W is the midpoint of QR. Point S liesˆ .on the x-axis and PRQP(5;2)SQ(1 ; 1)TW R(9; 5)3c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)QUESTION 33(a)3(b) 2 5 4 ( 1) E ; 2 2 7 3 E ; 2 2 7 3 E ; 2 2 7 3 x x y yC E ; A C ; A 2 2 2 2 7 3 1 xC 3 yC E ; ;2 2 2 27 1 xC3 3 yC or2222 7 1 xCor 3 3 yC xC 6or(2)7 1 xC 223 3 yC 22 xC 6 yC 0 C(6 ; 0) (5)yC 0 C(6 ; 0)3(c)4 3 1 12 1 10 ( 1) 1mCD 16 51 mAB mCDmAB AB CD3 ( 1) 4 1 41 54 0 4mBC 12 6 4 mAD mBCmAD mAB mCD AB CD mAD mBC AD BC ABCD is a parallelogram mAB mAD 1 Â 90 ABCD is a rectangle(10) AD BC ABCD is a parallelogramNow mAB mAD (1) ( 1) 1 AB AD Â 90 ABCD is a rectangle(since one interior angle of parallelogram ABCD is 90 )4c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)QUESTION 22(a)2(b)1 ( 3) 2 k ; 2 22 k 1;3 1; 2 2 k 3 2 6 2 k k 4 1;3 mAB mCD lines // 2 1 k ( 3) 1 3 3 21 k 3 2 5 5 2(k 3) 5 2k 6 2k 11 k 2mAB mCD 1 lines 2(c)1 k 3 1 2 5k 3 110 k 3 10 k 13mAB mBC lines // 2(d)11 k 2 3 ( 3)1 1 k 26 6 2(1 k ) 6 2 2k 8 2k k 4CD 5 222And CD2 2 ( 3) 3 k 2(e) 5 2 2 52 9 6k k 2 50 25 9 6k k 2 0 k 2 6k 16 0 (k 8)(k 2) k 8 or k 2 1 ( 3) 2 k 1;3 ; 2 22 k 3 2 k 4(3)2 1 k ( 3) 1 3 3 21 k 2 (3)1 k 3 1 2 5 k 13 (3)11 k 2 3 ( 3) k 4 (3) CD2 2 ( 3) 3 k 2 5 2 22 52 9 6k k 2 0 k 2 6k 16 0 (k 8)(k 2) k 8 or k 2(5)[17]5c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICS1(e)GRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15QT 2 (1 3) 2 ( 1 ( 2)) 2 QT 2 4 1 QT 2 5 QT 5TR 2 (3 9) 2 ( 2 ( 5)) 2 TR 2 36 9(LEARNER NOTES) correct substitution toget QT answer for QT correct substitution toget TR answer for TR establishing that1QT TR3(5) TR 2 45 TR 45 9 5 3 51 TR 5 QT31 QT TR31(f) tan tan 1 135 63, 43494882 ˆ 71,56505118 TPR 180 45 18, 43 tan mPR2 ( 5)5 9 tan 1(5) 135 tan mPT tan 2 63, 43494882 ˆ Now TPRˆ TPRˆ 135 63, 43494882 TPRˆ 71,56505118 TPR 90 71,56505118 180 18, 43 [24]6c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICS1(a)GRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15 1 (9) 1 ( 5) W ; 2 2 W(5 ; 3)(LEARNER NOTES) midpoint x 5(2)The equation of PW is x 51(b)1(c)1 5 ( 1) 4 9 1821(PS QR) mPS 21y 2 ( x 5)215 y 2 x 2219 y x 22 mQRmPT 2 mPT correct substitutioninto formula forequation y 2x 8mQR (PT QR)y 2 2( x 5) y 2 2 x 10 y 2x 81(d)mQR mPS correct substitutioninto formula forequation19 y x 22(4)(3) correct substitutioninto formula forequation11 y x 2211 x 2x 822 x 3 T(3; 2)121y ( 1) ( x 1)211 y 1 x 2211 y x 2211 x 2 x 822 x 1 4 x 16(5) 5 x 15 x 3 y 2(3) 8 2 T(3; 2)7c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)(a)(b)(c)(d)Determine the equation of PW if W is the midpoint of QR.Determine the equation of PS.Determine the equation of PT.Determine the coordinates of T.(e)Show that QT TR .(5)(f)Calculate the size of rounded off to two decimal places.(5)[24]13QUESTION 2:(2)(4)(3)(5)15 minutesConsider the following points on a Cartesian plane:A(1;2), B(3;1), C(-3;k) and D(2;-3)Determine the value(s) of k if:(a)(b)(c)(d)(e)( 1; 3) is the midpoint of AC.(3)(3)(3)(3)(5)[17]AB is parallel to CD.AB CD.A, B and C are collinear.CD 5 2QUESTION 3:25 minutesABCD is a quadrilateral with vertices A(1;3), B(2;4), C and D(5; 1) .The diagonals BD and AC bisect each other at point E.B(2;4)A(1;3)C D(5; 1)8c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)SESSION 20TOPIC: REVISION OF ANALYTICAL GEOMETRY (GRADE 11)Learner Note: Analytical Geometry is an important topic that carries a lot of marks in thematric final exam. Make sure that you know the basic formulae and then practise lots ofexamples involving applications of these formulae. The properties of quadrilaterals areextremely important in Analytical Geometry. Make sure you can prove that a quadrilateralis a parallelogram, rectangle, square, rhombus or trapezium by knowing the properties ofthese quadrilaterals.SECTION A: TYPICAL EXAM QUESTIONSQUESTION 1:15 minutesIn the diagram, PQRS is a trapezium with vertices P(5; 2), Q(1; 1), R(9; 5) and S , andPS//QR. PT is the perpendicular height of PQRS and W is the midpoint of QR. Point S liesˆ .on the x-axis and PRQP(5;2)SQ(1 ; 1)TW R(9; 5)9c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)Properties of QuadrilateralsParallelogram:Both pairs of opposite sides are parallel.Both pairs of opposite sides are equal in length.Both pairs of opposite angles are equal.Both diagonals bisect each other.Rectangle:Both pairs of opposite sides are parallel.Both pairs of opposite sides are equal in length.Both pairs of opposite angles are equal.Both diagonals bisect each other.Diagonals are equal in length.All angles are right angles.Rhombus:Both pairs of opposite sides are parallel.Both pairs of opposite sides are equal in length.Both pairs of opposite angles are equal.Both diagonals bisect each other.All sides are equal in length.The diagonals bisect each other at 90 .The diagonals bisect both pairs of opposite angles.10c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSSENIOR SECONDARY INTERVENTION PROGRAMMEGRADE 12SESSION 15(LEARNER NOTES)SECTION B: ADDITIONAL CONTENT NOTESREVISION OF ANALYTICAL GEOMETRY (GRADE 11)If AB is the line segment joining the points A( xA ; yA ) and B( xB ; yB ) , then thefollowing formulas apply to line segment AB.The Distance FormulaAB2 ( xB xA )2 ( yB yA )2or AB ( xB xA )2 ( yB yA )2The Midpoint Formula x xB yA yB M A; where M is the midpoint of AB.2 2 The Gradient of a line segment joining two pointsGradient of AB yB yAxB xAParallel linesParallel lines have equal gradients. If AB CD then mAB mCDPerpendicular linesThe product of the gradients of two perpendicular lines is 1 . If AB CD, thenmAB mCD 1The equation of the lineInclination of a lineory yA m x xA tan mABIf mAB 0, then is acuteIf mAB 0, then is obtuseCollinear points (A, B and C)Using the gradient formula:mAB mBCor mAC mABorUsing the distance formula:mAC mBC11c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICS(a)(b)(c)(d)(e)(b)SESSION 15(LEARNER NOTES)Determine the coordinates of E, the midpoint of BD.Determine the coordinates of C.Show that ABCD is a rectangle.Determine the area of ABCD.Calculate the size of the angle rounded off to the nearest degree.QUESTION 4:(a)GRADE 12SENIOR SECONDARY INTERVENTION PROGRAMME5 minutesDetermine the numerical value of p if the straight line defined by theequation px 3 y 6 has an angle of inclination of 135 with respect tothe positive x-axis.Calculate the value of k if the points A(6;5), B(3;2) and C(2 k ; k 4) arecollinear.QUESTION 5:(2)(5)(10)(6)(8)[31](4)(3)[7]15 minutesIn the diagram below, a(-4;5), C(-1;-4) and B94;1) are the vertices of a triangle in aCartesian plane. CE AB with E on AB. E is the midpoint of line CD. AF BC withF on CB. The equation of AF is.(a)(b)(c)(d)Determine the equation of CD.Determine the coordinates of E.Determine the equation of the line through D and parallel to line AC.Determine, showing all calculations, whether the x-intercept of line CDalso lies on the line through AF with the equation.(4)(5)(6)(3)[18]12c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)QUESTION 33(a)3(b) 2 5 4 ( 1) E ; 2 2 7 3 E ; 2 2 7 3 E ; 2 2 7 3 x x y yC E ; A C ; A 2 2 2 2 7 3 1 xC 3 yC E ; ;2 2 2 27 1 xC3 3 yC or2222 7 1 xCor 3 3 yC xC 6or(2)7 1 xC 223 3 yC 22 xC 6 yC 0 C(6 ; 0) (5)yC 0 C(6 ; 0)3(c)4 3 1 12 1 10 ( 1) 1mCD 16 51 mAB mCDmAB AB CD3 ( 1) 4 1 41 54 0 4mBC 12 6 4 mAD mBCmAD mAB mCD AB CD mAD mBC AD BC ABCD is a parallelogram mAB mAD 1 Â 90 ABCD is a rectangle(10) AD BC ABCD is a parallelogramNow mAB mAD (1) ( 1) 1 AB AD Â 90 ABCD is a rectangle(since one interior angle of parallelogram ABCD is 90 )13c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICS3(d)SENIOR SECONDARY INTERVENTION PROGRAMMEGRADE 12SESSION 15(LEARNER NOTES)AB2 (2 1) 2 (4 3) 2 AB2 (2 1)2 (4 3)2 AB2 1 1 AB 2 AD2 (5 1)2 ( 1 3)2 AB2 2 AD 32 Area ABCD (4 2)( 2) AB 2AD 2 (5 1) 2 ( 1 3) 2 Area ABCD 8 units 2 AD 16 162(6) AD 32 16 2 4 2Area ABCD AD AB Area ABCD (4 2)( 2) 8 units 2B(2;4)A(1;3) C(6;0) D(5; 1)3(e)tan mBD tan 4 ( 1)2 55 tan 3 180 59 121 tan 121 53 tan 1 45 18, 43 ˆ 45 OCD 121 45 76 (8)14c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 15(LEARNER NOTES)tan mDC0 ( 1) 1 16 51 tan 1 45 ˆ 45 OCD tan 121 45 76 [31]QUESTION 44(a)4(b)3 y px 6p y x 23p tan135 3p 1 3 p 3mAB mBCpx 23p tan135 3p 1 3 p 3 y (4) mAB mBC working out gradients k 52 5 k 4 2 3 62k 3k 2 1 2k 3 2k 3 k 2 k 5 (3)[7]15c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATIONMATHEMATICSGRADE 12SENIOR SECONDARY INTERVENTION PROGRAMMESESSION 20(LEARNER NOTES)QUESTION 55(a)5(b)Equation of ABEquation of CD correct substitution intoformula for equation of line (4) E(2;2)5(c)C(-1;-4)E(2;2)Dand(5) correct substitution intoformula for equation of line (6)NowEquation of line required:5(d) Substitute (1;0) LHS RHSx-intercept:(3)Substitute (1;0) intolies on the line AF[18]16c) Gauteng Department of Education, 2013

GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME MATHEMATICS GRADE 12 SESSION 15 (LEARNER NOTES) SESSION 15 TOPIC: REVISION OF ANALYTICAL GEOMETRY (GRADE 11) Learner Note: Analytical Geometry is an important topic that carries a lot of marks in the matric final exam.