GR 11 MATHS ANALYTICAL GEOMETRY

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1 GR MATHS ANALYTIAL GEMETRY Gr Maths Analtical Geometr hecklist: The Drawers of Tools onsider 'drawers' of tools - all BASI FATS. Use these to analse the sketches, to reason, calculate, prove.... Distance, Midpoint & Gradient Parallel lines, Perpendicular lines & ollinearit NB: Bear in mind ase, ase, and ase on page 5.6 & 5.7 AB D m AB m D For an two fied points, P( ; ) & Q( ; ) Q( ; ) - Note: Vertical length QR - Horizontal length PR - AB D m AB - m D A, B and are collinear points... m AB - m D m A B m A ; ma B also means: m AB % m D - m B ; ma mb P( ; ) θ R - θ is the angle of inclination of the line PQ tan θ opposite adjacent - - The Angle of Inclination of a line The ø of inclination of a line is the ø which the line makes with α β the positive direction of the -ais. Distance PQ... PQ ( - ) + ( - ) PQ ( ) + ( ) Gradient PQ... Midpoint of PQ... the sum of the squares! (Pthag.) NB: If α or β is the angle of inclination (measured in degrees), then the gradient of the line tan α or tan β (which is a ratio or number). Given α or β, one can find the gradient: r, given the gradient, one can find α or β:... a number... an angle (measured in degrees) m PQ - - ( tan θ) change in change in + + ; Average of the 's & of the 's opright The Answer

2 Gr Maths Analtical Geometr Equations of lines NTES NB: Bear in mind ase, ase, and ase on p Standard forms: General: m + c or - m( - ) m... when c 0... lines through the origin c... when m 0... lines -ais k... lines -ais Finding the equation of a line: Special focus through given points... find m first through point and or to a given line... substitute m and the point. the gradient m + c the point we need the gradient & a point the gradient - m( - ) the point Y-cuts and X-cuts: Put 0 and 0, respectivel. Point of intersection of graphs: Solve the equations of the graphs simultaneousl. If a point lies on a line, the equation is true for it, and, vice versa... If a point satisfies the equation of a line, the point lies on the line. e.g. If a line has the equation +, then all points on the line can be represented b (; + ). opright The Answer

3 ANALYTIAL GEMETRY QUESTINS Grade 0 & (essential for Grade!) FRMULAE (distance, midpoint, gradient), ø F INLINATIN & STRAIGHT LINES An etract from our Gr Maths in. A(-; ), B(7; ) and (; ) are points in a artesian plane. alculate if:. B units. the gradient of B is. (6)(). is the midpoint of AB.. B -ais ()(). P(; ), Q(; -) and R(8; -) and the origin, are points on the artesian plane. Write down the following: A SKETH IS ESSENTIAL!. (a) the length of P. (a) the midpoint of P (b) the length of PQ (b) the midpoint of PQ (c) the length of QR (c) the midpoint of QR. (a) the gradient of P. (a) the equation of P (b) the gradient of PQ (b) the equation of PQ (c) the gradient of QR (c) the equation of QR (). If M(; -) is the midpoint of PQ and the coordinates of point P are (; 8), then determine the coordinates of Q. (5). A(; 8) and B(- ; 6) are points in a artesian plane. Determine:.. the gradient of AB ().. the gradient of D if D AB ().. the gradient of MN if MN AB (). Prove that M(0; ), N(; -) and P(; - 5) lie on a straight line (are collinear). (). Draw simple sketches of the following graphs: (a) (b) (c) (d) - (e) + (f) - + (g) (h) + 6 (i) - 5 (8).. Show that ΔARM is right-angled. ().. Show that RÂM 5º; giving reasons. (5). alculate the area of ΔARM. () 5. P(; 7) and Q(; -) are two points in a artesian plane. Determine: 5. the length of PQ (leave the answer P(-; 7) M in simplified surd form). () θ 5. the coordinates of M, Q(; -) the midpoint of PQ. () 5. the equation of PQ, in the form... () 5. the size of θ, the angle between PQ and the positive -ais. () 5.5 the equation of the line which is parallel to PQ and passes through the point (-5; ). The equation must be in the form... () 6. Determine the angle that + 5 makes with the positive -ais. (Rounded off to one decimal digit.) () 6. Determine the numerical value of p if the straight line defined b p + has an angle of inclination 5º with respect to the positive -ais. () 7. In the figure, QRST is a R(9; 9) parallelogram with vertices Q Q and T ling on the -ais. The side RS is produced to W S U such that RS SU. The length of QT is units T(0; ) and the coordinates of R and T U are (9; 9) and (0; ) respectivel. The line segment QU intersects TS at W. 7. Determine the coordinates of: 7.. Q 7.. U ()() 7. Determine the equation of line R. () 7. Now, if W is the midpoint of UQ, determine whether W lies on line R (i.e. whether, W and R are collinear). () 8. In the figure AB is a triangle with vertices A(; 5), B(-; -) and. T(; ) is the point of intersection of the altitudes (perpendicular or heights) from A, B and. The inclination of AB to the -ais is θ. The equation of B(- ; -) the line passing through and T is given b A(- ; -), R(; ) and M(6; -) are the vertices of a triangle. 8. Determine the length of AB. (Leave the answer in simplest surd form.) ().. alculate the coordinates of S, the midpoint of AM. ().. Determine the equation of line RS. () 8. alculate θ. ().. alculate the length of RA. () 8. Write down the gradient of AS. () opright The Answer θ S A(; 5) T(; ) Gr Maths Analtical Geometr: Questions Learners must know the definitions and properties of geometric figures. Revise 'Quadrilaterals' at the back of the book. 8. Show that the equation of B is given b () 8.5 Hence, determine the coordinates of point. () 9. Determine the gradient of a straight line if it was 9. parallel to the line , or 9. perpendicular to the same line. () 0. A(; ), B(-; -) and (; ) are the vertices of triangle AB in a artesian plane. 0. Determine the coordinates of the midpoint D of AB. () 0. Find the equation of the perpendicular bisector of AB. () 0. If E is the midpoint of A, determine the equation of BE. () 0. Show that DE B. () 0.5 alculate the magnitude of AB. (5) 0.6 Determine the coordinates of M so that AMB, in this order, is a parallelogram. (). Triangle PRS has vertices P(-; ), R(; -) and S(a; b), as shown in P(-; ) the accompaning sketch.. Show that T(; ) is the midpoint of PR is. (). If the perpendicular bisector R(; -) of PR passes through S, S(a; b) show that a b. (). If a < 0, b < 0 and the area of ΔPRS square units, find the coordinates of S. (8). If Q(; ) lies on the perpendicular bisector, eplain wh PQRS is a rhombus. (). The vertices of ΔABE is A(0; ), B(5; ) and E(; ).. Prove that Ê 90º. (). If ABD is a rhombus with diagonals A and BD intersecting at E, determine the coordinates of and D. (). Prove that ABD is a square, giving reasons. (). Determine: P N. the size of α. () 6 θ. the size of β. () R. the size of θ. () M α - β Q

4 Gr 0 & (essential for Grade!) FRMULAE (distance, midpoint, gradient), ø F INLINATIN & STRAIGHT LINES. B ( - 7) + ( - ) â ANALYTIAL GEMETRY â â ( - 8)( - 6) 0 â 8 or 6 P(; ) Q(; -) opright The Answer (a) 5 units (b) units (c) units. (a) ; (b) (; ) (c) (6; -). (a) (b) undefined (c) 0. (a) (b) (c) -. B inspection : Q(; -) ANSWERS Notice : P is from the origin P & Q have the same -coordinates PQ -ais (or -ais) Q & R have the same -coordinates QR -ais (or -ais) + Q 8 + Q R: & - â + Q â 8 + Q - 6 R(8; -) â Q â Q - â Q(; -) You don t need an formulae!. m B â - 7 â 9 The translation P M is ( - ; - ) â M Q is the same Q(; -) 7 M(; - ) (7; ) B(7; ) P(; 8).. m AB m MN - 7. m MN â M, N & P are collinear A.. m D 7 - & m MP as well. (a) (b) (c) (d) (e) (f) (g) & (h) : There is no need to write these equations in standard form! (g) intercept : (when 0) - -intercept : (when 0) - Use the dual-intercept method (h) + 6 -intercept : (when 0) -intercept : (when 0) (i) - 5 -intercept : (when 0) - 5 -intercept : (when 0) S ; â S(; - ).. m RS - (- ) â Equation of RS : Substitute m 5 & point (; ) in m + c R - m( - ) â (5)() + c - 5( - ) â -7 c â â Eqn. : A(- ; -) - R(; ) S - -5 â 5-7 M(6; - ).. RA ( + ) + ( + ) â RA.. m AR - (-) - (-) - (- ) & m RM - 6 Gr Maths Analtical Geometr: Answers 5 j 7, units â m AR % m RM - â ARM 90º i.e. ΔARM is right-angled.. We have RA 5... in.. & ARM 90º... in.. Now, RM ( - 6) + ( + ) â RM 5 ( RA!! ) â ΔARM is an isosceles right-angled Δ â RAM 5º. Area AR.RM... bh square units 5. PQ ( + ) + ( - 7) â PQ units (-) 5. M ; 5. m PQ (- ) -8 - â M(; ) Subst. P(-; 7) & m - in m + c â 7 (- )(-) + c â 7 + c â 5 c â Equation of PQ : tan θ m PQ - â θ 80º - 6,º 6,6º R : - m( - )

5 5.5 Substitute m - & (- 5; ) in - m( - )... â - - ( + 5) â â â â â Gradient tan θ - â θ j 80º -,7º 6,º 6. p Q(0; 7) â â p (0; ) 7.. SU RS... given (QT )... opp. sides of m 0 p + â UR 8 units & UR -ais... opp. sides of m â U R 5º â U(9; ) The gradient of W j 0,9 or, m + c tan 5º - â p - whereas the gradient of R m R m W, W & R are not collinear Q (0; 7) T it goes through (0; 0) and (9; 9) i.e! or m 9 9 & c 0 W R(9; 9) S(9;...) U(9; ) 8. AB ( + ) + (5 + ) ( 7) â AB units 8. m AB 5 - (- ) - (- ) â θ 5º 8. m AS ( tan θ) 8. Substitute m B - & point B(- ; -) in m + c or - m( - ) â - - (- ) + c â + - ( + ) â - + c â + - â - c Equation of B is - - (% ) â â â - -, etc. 8.5 At, equation of B & equation of T â â 7 â & â (; -0) has -intercept, - & -intercept, - â Gradient R: Standard form: â â Gradient of parallel line - 0. D - ; 0. m AB - (- ) - (- ) Gr Maths Analtical Geometr: Answers 6 â m perpendicular bisector - â D - ; 0. Equation of BE : & m - E midpoint A E ( ; ) â m BE + + in - m( - ): â & (- ; - ) : â + ( + ) â 0. m DE & m B ( ) - (- ) - (- ) 6 â - + â DE B... equal gradients 0.5 m B β 6,57º & m AB α 6,º â 0.6 M(- 5 ; ) AB α - β 6,º - 6,57º 6,86º The translation, A is: (; ) ( - ; + ) A(; ) E (; ) â Also B M... opposite sides of a parallelogram are equal and parallel â M(- - ; - + ) D α B(-; -) β â W does not lie on R R, For W,... and â W does not lie on R... its coordinates don t satisf the equation of R opright The Answer 9. & Gradient of perpendicular line + A. Midpoint of PR is (- ) T ; â T(; ) P(-; ) S(a; b) Q(; ) T(; ) R(; -)

6 . m PR (-) - - â m perpendicular bisector Substitute m & point (; ) in - m( - ) â - ( - ) â At S, a b... S on line. Area of ΔPRS PR. ST Now, PR ( + ) + (- - ) & ST (a - ) + (a - )... (b a in.) (a - )... â.. (a - ) â â (a - ) (a - ). (a - )!! â (a - ) 9 â a - ± â a ± â a -... (ä a < 0) â S(-; -) (a b). m AE & m EB â m AE % m EB - â AE EB, i.e. Ê 90º. (; - )... E midpoint A & D(-; -)... E midpoint BD. We just need to prove angle 90º it s alread a rhombus! m DA - (-) (-) 5 & m AB â m DA % m AB - â DA AB, i.e. DAB 90º â ABD is a square... A(0; ). m MN +. m PQ - 6 -,5 â α 5º β 80º - 56,º,7º. θ MRQ... verticall opposite ø s β - α... eterior ø of Δ 78,7º D E(; ) A rhombus with one angle of 90º B(5; ) Gr Maths Analtical Geometr: Answers NTES. The other solution to a in. is a + (& b a ) â For Q(; ), also on the r bisector of PR (like S), ΔPRQ also has an area of square units. â TQ TS R: ST (- - ) + (- - ) 8 & TQ ( - ) + ( - ) 8 ST TQ R: The midpoint of SQ is ; â T(; ) â The diagonals bisect one another (â m ) & the do so perpendicularl â PQRS is a rhombus opright The Answer A