JEE STUDY PACKAGE MATHEMATICS

Transcription

1 JEE STUDY PACKAGE MATHEMATICS SAMPLE CHAPTERS Straigt Lines Differentiation

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3 CONTENTS CH. STRAIGHT LINES Brief Review of Geometr Straigt Line and Angles Parallel Straigt Lines Polgons Quadrilaterals Triangles Centres of Triangle Wat Is Co-Ordinate Or Analtical Geometr? 5 Co-Ordinate Aes 5 Distance Formula 5 Section Formula 6 Centres Connected wit A triangle 7 A Brief Introduction to Determinants 9 Area of Triangle Locus Equation of Straigt Line in General Form 4 Equation of Straigt Line in Various Forms. 6 Angle between Two Lines 9 Lines Equall Inclined To Given Line Concurrenc of Tree Lines Equation of Straigt Line in Parametric Form Equation of Straigt Line in Determinant Form 4 Position of a Point w.r.t. a Line 5 Perpendicular Distance of a Point from A Line 6 Angle Bisectors of Two Lines 7 Angle Bisectors in Triangle 0 Famil of Straigt Lines Pair of Lines Second Degree Homogeneous 4 Transformation Of ais 7 Second Degree Non Homogeneous 7 Concept of Homogenization 4 JEE Main Objective Questions 45 JEE Advanced Objective Questions I 49 JEE Advanced Objective Questions II 5 JEE Subjective Questions 59 JEE Main / AIEEE Arcive 6 IIT-JEE / JEE Advanced Arcive 64 Answers 65 CH. DIFFERENTIATION Definition of Derivative 67 Derivative b First Principle 67 Derivative of Standard Functions 68 Derivative of Implicit Functions 74 Logaritmic Differentiation 76 Parametric Differentiation 78 Derivative of One Function w.r.t. anoter 80 Successive Differentiation 80 Derivative of Inverse of a Function 8 Derivative of Functions Epressed In te Determinant Form 85 Derivatives of Inverse Trigonometric Functions 86 Deduction of Identities b Differentiating A Given Identit 88 Differentiation Using Substitution 88 JEE Main Objective Questions 90 JEE Advanced Objective Questions I 94 JEE Advanced Objective Questions II 97 JEE Subjective Questions 0 JEE Main / AIEEE Arcive 0 IIT-JEE / JEE Advanced Arcive 04 Answers 05

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5 STRAIGHT LINES BRIEF REVIEW OF GEOMETRY In tis section we are quickl going to review some geometrical figures ou ave studied till class 0 t. STRAIGHT LINE AND ANGLES Definition If two angles AOB and BOC ave a common verte O, a common arm OB and if te are situated on te opposite sides of teir common arm, ten te are called adjacent angles. Teorem If a straigt line stands on anoter straigt line ten te sum of te two adjacent angles is two rigt angles. Teorem If te sum of two adjacent angles AOC and COB wit common arm is two rigt angles, ten and are supplementar ras. Definition Two angles are verticall opposite angles if te sides of one of tem are supplementar ras of te sides of te oter. Teorem If two straigt lines intersect te verticall opposite angles so formed are equal. PARALLEL STRAIGHT LINES Definition Two straigt lines on a plane are parallel if te ave no common point. If two lines ave all te points common ten te two straigt lines coincide. Teorem 4 Suppose two straigt lines are cut b a transversal and an one of te following tree conditions old, namel,. a pair of alternate angles are equal,. a pair of corresponding angles are equal,. a pair of interior angles on te same side of te transversal add upto 80, ten te two straigt lines are parallel. Corollar If two straigt lines and m are bot perpendicular to te same straigt line p, ten. Teorem 5 If two parallel straigt lines are cut b a transversal ten. te alternate angles are equal. te corresponding angles are equal. te sum of te interior angles on te same side of te trans versal is equal to 80. Corollar If a straigt line is perpendicular to one of two or more parallel straigt lines, ten it is also perpendicular to te oter parallel straigt lines. POLYGONS Polgons ma be divided in two broad categories; conve and concave polgons. Definition 4 If we join an two consecutive vertices of a polgon b a straigt line and in eac case all te vertices lies on te same side of te straigt line ten te polgon is called a conve polgon. If te vertices lies on bot sides of an straigt line joining an two consecutives vertices ten te polgon is called a concave polgon. Conve Concave We are interested onl in conve polgons as far as JEE sllabus is concerned. QUADRILATERALS Definition 5 A four sided polgon is called a quadrilateral. All te quadrilaterals ma be divided in te following sceme. Parallelogram: A quadrilateral is a parallelogram if it satisf an of te following properties. i. Opposite sides are parallel. ii. iii. iv. Opposite side are of equal lengt. Opposite angles are equal. Diagonals bisect eac oter. It sould be noted tat if a quadrilateral possesses an of te above properties it automaticall possesses all te oter tree properties. So we can sa tat an parallelogram will posses all te above properties. Rectangle: A parallelogram wit equal diagonals. (Besides te four Properties mentioned above) Rombus: A parallelogram wit mutuall perpendicular diagonals. (Besides te four Properties mentioned above) Square: Ma be defined as a rectangle wit all sides equal. (Besides te four Properties mentioned above) O R

6 Ma be defined as a rombus wit equal diagonals(besides te four Properties mentioned above) Areas: Rectangle: If sides are a and b ten area a, b. Rombus: If diagonals are d and d ten area d d. Square: If side is a ten area a. Miscellaneous results on Parallelograms: Teorem 6 Te sum of te interior angles of an conve polgon aving n sides is (n 4) rigt angles. Corollar Te sum of te eterior angles of a conve polgon is 60 or four rigt angles. Teorem 7 Parallelograms on te same base and between te same parallel are equal in area. Corollar Te area of a parallelogram is equal to te area of te rectangle wose adjacent sides are equal to te base and te altitude of te parallelogram. In oter words, te area of a parallelogram is te product of its base and te altitude. Teorem 8 If a parallelogram and a triangle are on te same base and lie between te same parallels, ten te area of te triangle is alf tat of te parallelogram. Trapezium: A quadrilateral aving eactl one pair of opposite sides parallel is called a trapezium. In general te diagonals do not bisect te figure in two congruent triangles. Two special tpe of trapezium: Isosceles trapezium: Non parallel sides are equal in lengt and base angles are equal but not equal to 60. Equilateral Trapezium: Non parallel sides are equal and base angles are bot 60. Kite: A quadrilateral in wic diagonals are perpendicular and onl one diagonal divide, te quadrilateral in two congruent triangles is called a kite. Cclic Quadrilateral: A quadrilateral in wic opposite angles are supplementar is called cclic quadrilateral. Tese quadrilaterals can alwas be circumscribed b a circle. TRIANGLES Definitions wic must be known to te student: Rigt triangle, acute triangle, Equilateral triangle Scalene triangle, Isosceles triangle, Equilateral triangle Median, Altitude, Angle bisector, perpendicular bisector of side. Some teorems on general properties of triangles: Teorem 9 Te sum of te interior angles of a triangle is 80. Corollar An eterior angle of a triangle is equal to te sum of te interior opposite angles. Teorem 0 Te sum of an two angles of a triangle is less tan a straigt angle. Corollar An eterior angle of a triangle is greater tan an of te two non-adjacent interior angles. Corollar In an triangle ABC, at most one of te angles A, B, C can be obtuse. Teorem If two sides of a triangle are not equal, ten te greater side as te greater angle opposite to it. Teorem If two angles of a triangle are unequal, te greater angle as te greater side opposite to it. Teorem Te perpendicular bisectors of te tree sides of a triangle concur at a point. Teorem 4 Te bisectors of te tree angles of a triangle meet at a point. Teorem 5 Te tree medians of a triangle meet at a point and te point of concurrence trisects eac median. Teorem 6 Te altitudes of a triangle meet at a point. Corollar Triangles on equal bases and between te same parallels are equal in area. Teorem 7 If a straigt line is drawn parallel to one side of a triangle, ten it divides te oter two sides proportionall. Also, conversel, if a straigt line divides two sides of a triangle proportionall, ten it is parallel to te tird side. Similar Triangles: Definition: Two triangles are similar if te tree angles of te one are equal to te tree angles of te oter taken in order and te sides about te equal angles are proportional. Some Teorems on Similarit of triangles Teorem 8 If a straigt line XY parallel to BC meets te sides AB, AC of a triangle ABC at X and Y respectivel, ten AXY ABC. Teorem9 Equiangular triangles are similar triangles. Teorem 0 It two triangles ave tree sides of te one proportional to te tree sides of te oter, taken in order, ten te two triangles are similar. Teorem If two sides of te one triangle are respectivel proportional to two corresponding sides of te oter and if te included angles are equal, ten te two triangles are similar. Teorem In te two triangles ABC and ABC we ave A A. Ten teir areas are proportional to te rectangles contained b te sides containing A l and A. Teorem Te areas of two similar triangles are proportional to te squares on corresponding sides. Congruence of Triangles: If two triangles ave two sides of te one equal to two sides of

7 te oter, eac to eac, and also te angles contained b tose sides equal, ten te two triangles are congruent. Some Teorems on congruence of triangles Teorem 4 (ASA teorem): Two triangles are congruent if two angles and a side of one triangle are respectivel equal to two angles and te corresponding side of te oter. Teorem 5 If two sides of a triangle are equal ten te angles opposite to tese sides are equal. Teorem 6 (SSS Teorem): If te tree sides of one triangle are respectivel equal to te tree sides of anoter triangle, ten te two triangles are congruent. Teorem 7 (RHS Teorem): If in two rigt angled triangles, te potenuse and a side of one triangles are respectivel equal to te potenuse and a side of te oter, ten te two triangles are congruent. Teorem 8 : Te internal (or eternal) bisector of A of ABC divides te opposite side BC internall (or eternall) in te ratio of te sides AB and AC containing te A. Definition : If a line segment AB is divided internall and eternall in te same ratio at P and Q respectivel, ten AB is said to be divided armonicall at P and Q. Te points P and Q are called armonic conjugates wit respect to AB. Note : (i). If P and Q divide AB armonicall, ten AP, AB and AQ are in armonic progression, Incentre: Point of intersection of internal angle bisectors is called Incentre of te triangle. A circle can be drawn wit centre at tis point wic touces all te tree sides of tis triangle. Note te following properties Teorem 9 If AB c, AC b and BC a ten B C AI b c BI a c CI a b ; ;. ID a IE b IF c F A I D a E b BD c CE a AF b ; ; DC b EA c FB a Ecentre: Te eternal bisectors of an two angles of a triangle and te internal bisector of te tird angle are concurrent and tat te point of concurrence is equidistant from all te tree sides of te triangle. Tese points are called Ecentres. C i.e.,. AB AP AQ A (ii). If O is te mid point of AB and P,Q divide AB armonicall OB ten OP. OQ. Centroid of triangle: CENTRES OF TRIANGLE Te point of intersection of medians of a triangle is called centroid of te triangle. Teorem 8 Centroid divides te median in te ratio : (verte to side) F B D r I C E A I A I G B I C B C M I

8 Ortocenter: Point of intersection of altitudes of a triangle is called ortocenter of te triangle. For a rigt triangle ortocenter is at te point te verte wit rigt angle. te triangle obtained b joining te feet of altitudes is called Pedal Triangle. Circumcentre: Point of intersection of perpendicular bisector of sides is called circumcentre of te triangle. A circle can be drawn wit circumcentre as te centre wic passes troug te vertices of te triangle.. Historical Development: EXAMPLE P is a point inside te triangle ABC. Lines are drawn troug P, parallel to te sides of te triangle. Te tree resulting triangles wit te verte at P ave areas 4, 9 and 49 sq. units. Find te area of te triangle ABC. \ similarl 49 q a p a or...() and 7 q a r a...()...() EXAMPLE Find te number of straigt lines equidistant from tree non collinear points in te plane of te points. Tree non collinear points form a triangle and te line joining te mid points of an two sides is equidistant from all te tree vertices. () + () + () gives p q r a D 44 Hence Lines EXAMPLE Te distance between te two parallel lines is unit. A point 'A' is cosen to lie between te lines at a distance 'd' from one of tem. Triangle ABC is equilateral wit B on one line and C on te oter parallel line. Find te lengt of te side of te equilateral triangle. cos ( + 0º) d () and sin - d () dividing cot d d, squaring equation () and putting te value of cot, (4d - 4d + 4) d d EXAMPLE 4 Let P is an point inside te triangle ABC of side lengts 6, 5, 5 units and p, p, p be te lengts of perpendiculars drawn from P to te sides of triangle. Find te maimum value of p.p.p. Area ABC Area BCP + Area CPA + Area APB p + 5 p + p Appling AM G.M 5 5 p p p 5 5 p p p 75 p p p 4 56 p pp 75 EXAMPLE 5 / If one side of a rombus as end points (4, 5) and (, ) ten find te maimum area of rombus. AB is given and O is intersection of diagonals. Let AO a and OB b ten a + b (4 ) + (5 ) 5. Area of rombus ab. / 4

9 Let a 5 sin, b 5 cos Ten ab 5 sin. So maimum area 5 wen 45. WHAT IS CO-ORDINATE OR ANALYTICAL GEOMETRY? Historicall geometr and algebra were separate brances of matematical ad none of tese correlation. Algebra was concerned wit solving equation, finding roots etc. and geometr was concerned wit geometrical sapes straigt lines, triangle, circle etc. and te properties associated wit tem. Te idea of unifing geometr and algebra emerged wit an attempt to locate te position of point in plane. Hipparcur (around 50 B.C.) used coordinates in astronom and geograp calling tem longitudes and latitudes. Oreseme (around 8) gave te idea of fiing up coordinate aes. But te full integration of geometr and algebra begin in 7 t centur wit te independent works of Fermatt (69) and Descartes (67). Now co-ordinate geometr is te full-fledged branc of matematics wic full unifies algebra and geometr. Using analtical geometr algebra in used to stud geometrical. Curves and tere properties in details and wit muc ease. In tis capter we will learn to use algebra for stud of geometrical sapes and properties associated wit tem wic were till now done using pure geometr onl. CO-ORDINATE AXES Te idea of using algebra in geometr begin wit te concept of coordinate aes We draw two mutuall perpendicular lines XX and YY intersecting at O in a plane. On OX and OY we assign eac point a +ve number, to eac point, increasing in OX and OY direction wit zero at O. Similarl negative real number are assigned on OY and OX Te line X OX is called - ais and te line Y OY is called -ais. If for a point P distance from -ais (PM) is and distance from ais is (PN) ten is called -coordinate or abscissa of P and is called coordinate or ordinate of te point P. You sould note tat ever point ling on te plane can be assigned coordinates like tis and coordinates of eac point are unique. X' N Y 4 O Y' P M X Note: A point wose abscissa and ordinate bot are integers are called Lattice Point (w.r.t. co-ordinate geometr). DISTANCE FORMULA Let P(, ) and P(, ) are two points in a plane ten as sown in figure. P (, ) QN PN B Ptagoras teorem for triangle PQN PQ PN QN Q (, ) N PQ ( ) ( ) Hence distance d between two points P(, ) and Q(, ) is given b d ( ) ( ) EXAMPLE 6 If P is a moving point in te plane in suc a wa tat perimeter of triangle PQR is 6 units, were Q is (, 5 ) & R is (7, 5 ) ten maimum area of triangle PQR is? Distance between Q & R is 6 units. Now triangle PQR will be of maimum area wen PQ PR 5 implies altitude will be 5 4. Hence area 6 4 square units. EXAMPLE 7 If M is te mid-point of te side BC of te triangle ABC, prove tat AB + AC AM + BM. Tis is a well-known teorem in pure geometr ; we prove it b analtical metods. In fig. coose BC to define te positive direction of te -ais and M to be te origin. Let te abscissa of C be a ; ten, since M is te mid-point of BC, te abscissa of B is a; tus C is (a, 0) and B is ( a, 0). Let te 5

10 coordinates of A be (, ). Ten AB [ ( a)] + [ 0] ( + a) +...(i) and AC ( a) +....(ii) Add (i) and (ii) and simplif, ten Hence coordinates of a point R dividing P & Q in te ratio m: n internall are m n m n & m n m n Similarl it can be proved tat coordinates of a point R dividing P & Q in te ratio m: n eternall are m n m n m n & m n Te coordinates of te mid-point of te line-segment joining AB + AC + a +....(iii) AM + and BM a, so tat AM + BM + a +....(iv) Te formulae (iii) and (iv) given te desired result, namel, AB + AC AM + BM. SECTION FORMULA Let P(, ) & Q(, ) are two point in a plane and point R divides PQ internall in te ratio m:n, i.e. PR : RQ m : n. As sown in figure. PS PT B similar triangle PQS ~ PRT. QS PQ RT PR m n m (, ) and (, ) are, Remark: If te ratio, in wic a given line segment is divided, is to be determined, ten sometimes, for convenience (instead of taking te ratio m : n), we take te ratio : and appl te formula for internal division. If te value of turns out to be positive, it is an internal division oterwise it is an eternal division. EXAMPLE 8 Two points A(, ) and B(, ) are cosen on te grap of f () ln wit 0 < <. Te points C and D trisect line segment AB wit AC < CB. Troug C a orizontal line is drawn to cut te curve at E(, ). If and 000 ten find te value of. Using section formula 000 a 4 (, ) m P (, ) (, ) n Q R T S b 0 ln000 b ln000...() m m (m n) m n m m m n Similarl, PS PQ PT PR n m m n m n m now line b intersects te curve ln b ln...() from () and () ln000 ln ln (000) / ln (000) / 0 Ans. 6

11 CENTRES CONNECTED WITH A TRIANGLE (w.r.t. ABC, were A (, ), B (, ), C (, ), BC a, CA b & AB c). Centroid : Te point of concurrenc of te medians of a triangle is called te centroid of te triangle. Te centroid of a triangle divides eac median in te ratio :. Te coordinates of centroid are given b Let A(, ) B(, ) C(, ) form a triangle. Let d is te mid point of BC ten te centroid G is te point dividing A and D in te ratio :. Coordinates of D,.... Coordinates of G, + + Hence centroid of ABC is,. Important Note Internal and eternal angle bisector of an angle of a divide te base Harmonicall. b DC Proof :In ADC, sin A sin c BD In ABD, sin( ) A sin...()...() G Ortocentre :, Te point of concurrenc of te altitudes of a triangle is called te ortocentre of te triangle. Note tat te distances of te ortocentre from te vertices of te ABC are R cosa, R cosb, R cosc ; and from te sides are. R cosb cosc, R cosc cosa, R cosa cosb. Te co-ordinates of te ortocentre are given b tana tanb tanc, tana tanb tanc tana tanb tanc tana tanb tanc BD c...() DC b AE is te eternal angle bisector. BE Again, in ABC, A sin CE In ACE, A sin Tus c b sin b c sin...(4)...(5) B E...(6) C E From () and (6) if follows tat internal and eternal angle bisectors divide te opposite base internall and eternall in te same ratio. Two suc points like D and E are known as armonic conjugate and it can be sown tat BD, BC and BE are in H.P. 7

12 Now BD BE DC CE BD BE BC BD BE BC BC BC BD BE BC BC BD BD BE BC BE B c L A b C Incentre : Te point of concurrenc of te internal bisectors of te angles of a triangle is called te incentre of te triangle. Te coordinates of te incentre are given b A(, ), B(, ) and C(, ) form a triangle. AD is te angle bisector of ÐA BD:DCc:b. AI : ID (b c):a c b c b D, b c b c c b c b (b c). a (b c). a I b c, b c a (b c) a (b c) Te Incentre of triangle ABC is a b c a b c, a b c a b c BL LC I c,also b AI b c I L a a b c a b c I, a b c a b c Note tat I & I ; I & I and I & I are armonic conjugate of eac oter. Circumcentre : Te point of concurrenc of te perpendicular bisectors of te sides of a triangle is called circumcentre of te triangle. Te coordinates of te circumcentre are given b sina sinb sinc sina sinb sinc,. sina sinb sinc sina sinb sinc A(, ) c B (, ) I D a b C (, ) a b c a b c I,. a b c a b c Ecentre: Co-ordinate of ecentre opposite to A is given b a b c a b c I, a b c a b c and similarl for ecentres (I & I ) opposite to B and C are given b a b c a b c I, a b c a b c ou are advised to derive te result in all above cases namel centroid, ortocentre, incentre, ecentre, circumcentre. Rater tan te end result, process is more important. For tis ver purpose, self eplained diagrams ave been given. Remarks :. Circumcentre O, Centroid G and Ortocentre H of a ABC are collinear. G Divides OH in te ratio :, i.e. OG : GH : Proof: O G C collinear Ds CGD & AGO are similar But CD (distance of c.c. from BC) R cosa OA distance of ortocenter from verte A R cosa AO C' D AG GD OG GO' 8

13 8 G, 9 A (, 0 ) AD : DG : G must be te centroid. Note in an isosceles D O; G; C and I are collinear. In an isosceles triangle centroid, ortocenter, incentre and circumcentre lie on te same line and in an equilateral triangle all tese four points coincide. EXAMPLE 9 Te co-ordinates of middle points of te sides of a triangle are (4, ) (, ) and (, ). Ten find te co-ordinates of its centroid. Centroid EXAMPLE 0, 7,. Find te coordinates of te incentre of te triangle formed b te lines and is Incentre a b c a b c, (0, - a b c a b c ) Here a, b, c. EXAMPLE If in triangle ABC, A (, 0), circumcentre, and ortocentre, 4 ten find te co-ordinates of mid-point of side opposite to A. Circumcentre ortocentre O is 4,, H is, 8 coordinates of G are, 9 A(, 0), D D 8 0 G (, /8 ) B D C coordinate of te mid point is BC are, EXAMPLE In a triangle ABC, A (, ), B (,), C (,) and point A lies on te line, were, I. If te area of triangle ABC be suc tat [ ], were [.] denotes te greatest integer function, ten te number of all possible coordinates of A must be? (, ) lies on Ten, Tus, te coordinates of A are (, )... (i) ( ) but [ ] Hence, possible coordinates of A are (, 7), (, 9), ( 7, ) and ( 6, 9) I,, 7, 6 Number of all possible coordinates of A are 4. A BRIEF INTRODUCTION TO DETERMI- NANTS An arrangement of numbers in rows and columns (n n) is called a determinant of order n. a b P a b is a determinant of order. 9

14 a b c Q a b c a b c Metod of evaluation: P a b b a is a determinant of order. Q a (bc cb ) b (ac ac ) c (ab ba ) Wen epanded against first row. (Sarrus Rule) Write te given determinant as follows : a a a b b b c c c a a a (abc bca ca b ) (a bc bca ca b ) PROPERTIES OF DETERMINANTS:. Reflection Propert Te determinant remains unaltered if its rows are canged into columns and te columns into rows.. All-zero Propert If all te elements of a row (column) are zero, ten te determinant is zero. b b b. Proportionall (Repetition] Propert If te elements of a row (column) are proportional [identical] to te element of te some oter row (column), ten te determinant is zero. 4. Switcing Propert Te intercange of an two rows (columns) of te determinant canges its sign. 5. Scalar Multiple Propert If all te elements of a row (column) of a determinant are multiplied b a non-zero constant, ten te determinant gets multiplied b te same constant. 6. Sum Propert a b c d a c d b c d a b c d a c d b c d a b c d a c d b c d 7. Propert of Invariance a b c a b c b c a b c a b c b c a b c a b c b c Tat is, a determinant remains unaltered under an operation of te form Ci Ci C j Ck, were j, k i, or an operation of te form Ri Ri R j R k, were j, k i. 8. Factor Propert If a determinant D becomes zero wen we put, ten ( ) is a factor of D. 9. Triangle Propert If all te elements of a determinant above or below te main diagonal consist of zeros, ten te determinant is equal to te product of te diagonal elements. Tat is, EXAMPLE a a a a b b a b 0 abc. 0 0 c a b c Witout epanding find te value of determinant Let ten appling C C ( 8)C ; 9 5 we get ( because two columns C and C are identical) EXAMPLE find relation in a, b, c wic are real if te value of determinant a ab ac b ab bc ac bc c Multipl R b a, R b b & R b c & divide te determinant b abc. Now take a, b & c common from c, c & c. Now use C C + C + C to get (a + b + c + ) b b b. c c c Now use c c c & c c c we get + a + b + c a b c 0 0

15 AREA OF TRIANGLE Let ABC be te triangle wit te coordinates of its vertices A, B and C being (, ), (, ) and (, ). We draw AL, BM and CN perpendicular to te ais of, and let D denote te required area. Ten D trapezium ALNC + trapezium CNMB trapezium ALMB LN(LA NC) NM(NC MB) LM(LA MB), [( )( ) ( )( ) ( )( )] On simplifing we easil ave ( ), Y O A or te equivalent form L C N [ ( ) ( ) ( )] If we use te determinant notation tis ma be written as,,,,,, Area of a quadrilateral: Let te vertices of te quadrilateral, taken in order, be A, B, C and D M Let teir coordinates be respectivel (, ), (, ), (, ) and ( 4, 4). Y A D C B B X LR(LA RD) RN(RD NC) NM(NC MB) LM(LA MB) [( )( ) ( )( ) ( )( )] {( ) ( ) ( 4 4 ) ( 4 4 )}. Remarks : In case of polgon wit vertices (, ), (, ),... ( n, n ) in order, ten area of polgon is given b ( ) + ( ) ( n n n n) ) + ( n n ) Collinearit of tree points To prove tree points P, Q, R collinear. We can use an of te following properties of tree collinear point. i. Area of te triangle formed b te tree points.p(, ), Q(, ) and R(, ) is equal to zero. Tus, if P, Q, R are collinear tan we must ave 0. ii. Slope of line joining an pair of points is equal. EXAMPLE 5 In an acute triangle ABC, if te coordinates of ortocentre 'H' are (4, b), centroid 'G' are (b, b 8) and circumcentre 'S' are ( 4, 8), ten find te possible value(s) of 'b'. As H, G and C are collinear 4 b 4 b b b b 4 b (b 4) 6 b 0 (b 4)(6 b) + (b + 4)(b 8) 0 (b 4)(8 b) + (b + 4)(b 8) 0 (8 b)[(b 8) (b + 4)] 0 (8 b)(b ) 0 Hence b 8 or, wic is wrong because collinearit does not eplain centroid, ortocentre and circumcentre. O L R N M We draw AL, BM, CN and DR perpendicular to te ais of. Ten te area of te quadrilateral X trapezium ALRD + trapezium DRNC + trapezium CNMB trapezium ALMB Now 8 4 b b 4

16 and 6 b b 8 b 8 But no common value of 'b' is possible LOCUS Wen a point moves in a plane under certain geometrical conditions, te point traces out a pat. Tis pat of te moving point is called its locus. For eample: i. Wen a point moves suc tat its distance from a fied point P is alwas constant equal to r, it traces a circle wit centre at P and radius equal to r. ii. Wen a point moves suc tat its distances from two fied point A & B are alwas equal it traces te perpendicular bisector of te line segment joining A and B. A P(, k) B iii. Wen a point moves suc tat sum of its distances form two fied points is constant te point traces out a figure called ellipse (-a, 0) F (-c, 0) F F (c, 0) (a, 0) iv. Point A moves so tat Ar. DPAB ab Locus of P is a line to -ais/base of te D. Equation of locus Te equation to a locus is te relation wic eists between te coordinates of an point on te pat, and wic olds for no oter point ecept tose ling on te pat. In oter words equation to a curve (or locus) is merel te equation connecting te and te coordinates of ever point on te curve. Procedure for finding te equation of te locus of a point : i. If we are finding te equation of te locus of a point P, assign coordinates (, k) or (, ) to P. ii. Tr to write te geometrical restrictions imposed on te moving point P in form of algebraic equations and from tese tr to get a relation in & k involving onl te given quantities. ii. If and k coordinates of te moving point are obtained in terms of a tird variable t (let) called te parameter, eliminate t to obtain te relation in and k and simplif tis relation. iv. Replace b, and k b, in te eliminant. Te resulting equation would be te equation of te locus of P. Remember tat tere is no routine procedure to obtain equation of a given locus. Given above are just guidelines. Several oter metods will appear. EXAMPLE 6 Find te equation of locus of a point wic moves suc tat te sum of its distances from (, 0) and (-, 0) is 8 Step -Let te moving point be, k Step -Here, known quantities are (,0) and (-,0) Step - Sum of (, k) from (,0) and (-,0) is 8 k k k k 4 4 k k 6 k 8 k k 6 64 k 4k 48 6 Step 4- Here, tere is no unknown quantit to be eliminated Step 5- Substitute and k for and. Hence, is te desired equation of locus 6 EXAMPLE 7 Let A(5, ), B( cos, sin) and C( sin, cos) are angular points of ABC were R. Find te locus of ortocentre of ABC Clearl circumcentre is (0, 0) Now centroid G is 5 cos sin sin cos, G divides HO internall : and 5 cos sin (0) () sin cos (0) (k) H (ortocentre) : G (centroid) O (circumcentre)

17 sin cos and sin cos 5 k From () and (), we get k k... ()... () Locus of ortocentre H(, k) is Ans. EXAMPLE 8 A variable line is drawn troug te origin O. Two points A and B are taken on te line suc tat OA and OB units. Troug points A and B two lines are drawn making equal angle ' α ' wit te line AB. Find te locus of te point of intersection of te lines. Let C be te mid point of AB. Given tat AB AC CB OC and RC tanα OR OC RC tan α. 4 9 tan α is te required locus of te point of 4 intersection of te lines. STRAIGHT LINE Definition: Straigt line is defined as te locus of a point suc tat if an two points of tis locus are joined, te define a unique direction. To understand te concept of direction, we sould know inclination of a line and slope of a line. Inclination of a line: If a straigt line intersects te -ais, te inclination of te line is defined as te measure of te smallest non-negative angle wic te line makes wit te positive direction of te -ais. X' Y Y' Slope (or Gradient) of a line : 80 - If te inclination of a line (i.e. non vertical line) is and X, Ten te slope of a line is defined to be tan. Slope formula: If is te angle between at wic a straigt line is inclined to te positive direction of -ais, and 0 80, 90 0,, ten te slope of te line, denoted b m, is defined b m tan. If 90, m does not eist, but te line is parallel to te -ais. If 0, ten m 0 and te line is parallel to te -ais (i.e. line as zero slope). Y 0 o Y o X X Y Y 5 o 90 o o m tan0 / m tan5 / o m tan 0 0 m tan90 If A(, ) and B(, ),, are points on a straigt line, Ten te slope m of te line is given b o X X m (Since slope of a non-vertical line is te slope of an segment contained in te line). Note: (proof is given later) If two lines wit slopes m & m are Parallel ten m m and if te are perpendicular ten m.m - EXAMPLE 9 A rectangular billiard table as vertices at P(0, 0), Q(0, 7), R(0, 7) and S (0, 0). A small billiard ball starts at M(, 4) and moves in a straigt line to te top of te table, bounces to te rigt side of te table, ten comes to rest at N(7, ). Find te -coordinate of te point were it its te rigt side. a 7 a tan 0 b ence b a 7 a 0 b 7 4 also tan b b.

18 from st two relations 9 ab b a a + 6 ab b...() from last two 0a ab 0 + b a a ab + b or ab b a...() ence from () and () a + 6 a 0a 7 a.7 Ans. EXAMPLE 0 Vertices of a parallelogram ABCD are A(, ), B(, 6), C(, ) and D(, 6). If a line passing troug te origin divides te parallelogram into two congruent parts ten find te slope of te line. as sown m ; 5 Hence m 8 Ans. 4 8 EXAMPLE Given A(0, 0) and B(, ) wit (0, ) and > 0. Let te slope of te line AB equals m. Point C lies on te line suc tat te slope of BC equals m were 0 < m < m. If te area of te triangle ABC can be epressed as (m m ) f (), ten find te largest possible value of f (). Let te coordinates of C be (, c) c m ; m m m c m c m (m m ) c m c (m m ) + m...() now area of ABC [c m ] 0 0 m c [( m m ) m m ] [(( m m) m) m ] (m m )( ) ( > in (0, ) Hence, f () ( ); f ()] ma 8 wen EQUATION OF STRAIGHT LINE IN GENERAL FORM In tis section we will sow tat an straigt line can be written as a linear equation and converse of it. Statement: te general equation of first degree in and alwas represents a straigt line. Proof: Let te general equation of first degree in and be a b c 0 (i) Let A(, ) and B(, ) be an two points on locus (), ten a b c 0 (ii) and a b c 0 (iii) multipling (iii) b m and (ii) b n and adding, a(m n ) b(m n ) c(m n) 0...(iv) m n m n a b c 0 m n m n...(iv) m n m n Equation (iv) sows tat te point, m n m n i.e. te point wic divides AB in te ratio m: n lies on locus (i). 4

19 Since m and n are arbitrar numbers, terefore, eac point on line AB will lie on locus (i). Hence equation (i) represents a straigt line. Alternate proof: Let te general equation of first degree in and be a b c 0 (i) Let A(, ), B(, ) and C(, ) be an two points on locus (i), ten a b c 0 (ii) a b c 0.. (iii) and a b c 0.. (iv) Eliminating a, b, c from equations (ii), (iii) and (iv), we ave 0, terefore points A, B, C are collinear. Hence equation (i) represents a straigt line. Statement: Ever straigt line ma be represented b a first degree equation in and. (converse) Proof: We know tat troug two given points one and onl one straigt line can be drawn. Let A(, ) and B(, ) be two given points on te straigt line and let P(, ) be an point on it. Now since points P, A, B are collinear. 0 or ( ) ( ) 0 wic is of te form a b c 0. (i) were a, b and c. Since equation (i) is satisfied b all points on line AB and it will not be satisfied b te co-ordinates of an point wic does not lie on line AB, ence (i) is te equation of line AB. Hence equation of a straigt line can be taken as a b c 0, were a, b, c are constants. Note: Altoug, tere are tree constants a, b, c in te equation a b c 0 of a straigt line, but we can divide bot sides of tis equation b an one of tese constants wic is not zero and get te equation wic will contain onl two constants. For eample, a b c 0, were c 0 can be written as a b 0 c c A B 0, were a b A and B. c c Hence in order to completel determine te equation of a straigt line we require onl two conditions. Point of intersection of two lines Let te equation of te two given lines be a b c 0. (i) and a b c 0.(ii) Let (a, b) be te point of intersection of lines (i) and (ii), ten a b c 0 (iii) and a b c 0 (iv) Solving equations (iii) and (iv) b cross multiplication metod we ave b c b c c a c a a b a b b c from (i) and (iii), a b c a From (ii) and (iii), a b (i) (ii) (iii) b c a b c a. a b Hence te co-ordinates of te point of intersection will bc bc ca ca be, ab ab ab ab Note: (i) Te co-ordinates of te point of intersection of two lines can be obtained b solving simultaneousl te equations of te two lines. (ii) If te two lines are parallel, ten a b a b 0 a a b or b a and b are nor defined i.e. point of intersection do not eist. In tis case neiter ca ac nor bc bc will be zero, oterwise (same). a b c i.e. te two lines will be coincident a b c 5

20 EXAMPLE Find te area of te triangular region in te first quadrant bounded on te left b te -ais, bounded above b te line and bounded below b te line 5 +. Clearl lines intersect at P(0, 7) Y (0, 4) (0, 40 ) O P(0, 7) (4,0) (4, 0) 5 5 Area of saded region (square units) EXAMPLE X 50 In a triangle ABC, coordinates of A are (, ) and te equations of te medians troug B and C are + 5 and 4 respectivel. Find te coordinates of B and C. Let B ( b, b ) and C ( c, c ) and medians troug B and C be BB and CC respectivel. Clearl midpoints of line segments AB and AC would lie on te lines 4 and + 5 respectivel. b + 8, c c 5 b 7, c + c 7 If G be te centriod we ave G (4, ) 4 Also, b c b c c b 4 c 7 c Tus B (7, -) and C (4, ). b + c b - c - EQUATION OF STRAIGHT LINE IN VARIOUS FORMS.. Equation of Straigt Line in Slope intercept form Intercept on and ais: Te lengts cut off b te straigt line from -ais and -ais (between origin and te point of intersections) are called intercepts on and ais respectivel. If te line cuts positive directions of aes ten intercepts are measured positive and if it cuts negative directions ten intercepts are measured negative. If te line is parallel to te ais ten intercept is considered infinite. If te line is parallel to te ais ten intercept is considered infinite Derivation: Te equation of a straigt line wic cuts off a given intercept c on te ais of and is inclined at a given angle a to te ais of L' Y C O Let C be a point on te ais of suc tat OC is c. Troug C draw a straigt line LCL inclined at an angle (tan m) to te ais of, so tat tan m. Te straigt line LCL is terefore te straigt line required We ave to find te relation between te coordinates of an point P ling on it Draw PM perpendicular to OX wic meets in N a line troug C parallel to OX. Let te coordinates of P be and, so tat OM and MP. Ten MP NP MN CN tan OC m. c, m c. Tis relation being true for an point on te given straigt line tus it is te equation to te straigt line. Note: To reduce te equation A B C 0 to te form m c. Given equation is A B C 0 B A C wic is of te form m c, A C were m and c. B B N P M L A C B B Hence Slope of te line A B C 0 is coefficient of coefficient of. Intercept of te line on -ais EXAMPLE 4 X cons tan t term. coefficient of If te intercept made on te line m b te lines and 5 is less ten 5, ten find te range of m. 6

21 Te distance between (, m) and ( 5, 5m) is less ten 5. ( 5 ) +( 5m m) < 5 9m < 6 4 m < 4 4 m.. Equation of Straigt Line in Double intercept form Te equation of te straigt line wic cuts off given intercepts a and b from te aes. Derivation: Let A and B be on OX and OY respectivel, and be suc tat OA a and OB b. Line joining AB is te required line. If P be an point (, ) on tis straigt line. Draw PM perpendicular to OX. An point ling on te line will alwas satisf following geometrical relation. OM PB and OA AB MP AP OB AB OM MP PB AP, OA OB AB i.e. Y B O. a b P M A Tis is terefore te required equation as it is te relation tat olds between te coordinates of an point ling on te given straigt line Note: To reduce te equation A B C 0 to te form, remember tat tis reduction is possible onl wen a b C 0. Given equation is A B C 0 A B C or C C, A B X A B C C, were C 0 Wic is of te form C C, were a and b. a b A B Hence Intercept of te line A B C 0 on C cons tan t term -ais A coefficient of C cons tan t term -ais B coefficient of Tus intercept of a straigt line on -ais can be found b putting 0 in te equation of te line and ten finding are value of. Similarl intercept on -ais can be found b putting 0 in te equation of te line and ten finding te value of.. Equation of Straigt Line in Perpendicular form Te equation of a straigt line in terms of te perpendicular distance from origin from te origin and te angle tat tis perpendicular makes wit te ais of Derivation As sown in te figure Y B O L N R P M A OR is te perpendicular from P and let its lengt be p. is te angle tat OR makes wit OX. P is an point, wose coordinates are and, ling on AB We draw te ordinate PM and also ML perpendicular to OR and PN perpendicular to ML. Ten OL OM cos. (i) and LR NP MPsin NMP. But NMP 90 NMO MOL. LR MPsin (ii) Hence, adding (i) and (ii), we ave X OM cos MPsin OL LR OR p, i.e. cos sin p. (ii) Tis is terefore te required equation as it is te relation tat olds between te coordinates of an point ling on te given straigt line. Note: (i) In normal form of equation of a straigt line p is alwas taken as positive and is measured from positive direction of -ais in 7

22 anticlockwise direction between 0 and. (ii) To reduce te equation A B C 0 to te form cos sin p Given equation is A B C 0, or A B C. (i) Case I : Wen C < 0 i.e. C 0 dividing bot sides of equation (i) b A B, we ave A B C A B A B A B Wic is of te form cos + sin p, EXAMPLE 5 Wat is te -intercept of te line tat is parallel to, and wic bisects te area of a rectangle wit corners at (0, 0), (4, 0), (4, ) and (0, )? Area of OADE Area of ABCD since Area of ADD' Area of A'DA so area of rectangle AA'OE Area of rectangle D'DCB BD' OA () A B Were cos, sin A B A B C and p. A B Case II. Wen C > 0 i.e. C 0 ; from (i), A B C A B C A B A B A B Wic is of te form cos + sin p, were and p cos A C A B, sin A B A B B. 4. Equation of Straigt Line in Point-Slope form Te equation of te straigt line wic passes troug a given points ( ', ') ave slope equal to m. Derivation: As te equation to an straigt line is m c. (i) B properl determining te quantities m and c we can make (i) represent an straigt line we want. If (i) pass troug te point ( ', '), we ave ' m ' c (ii) Subtracting for c from (ii), te equation (i) becomes ' m( ')... (iii) Tis is te required equation as it is te relation tat olds between te coordinates of an point ling on te given straigt line. m AD 0...() from () and () Equation of line AD is 0 put 0, get 5 Hence (0, 5) Ans. EXAMPLE 6 Find te point of intersection of te lines and and cuts off equal intercept on coordinate aes. Point of intersection of given lines are and ence equation of line is +, 8 8 8

23 EXAMPLE 7 Prove tat te area of te triangle formed b te lines m +c (c c) and m +c and -ais is given b. Hence m m deduce tat te area of te triangle formed b tree lines m +c ; m +c and m + c is (c c) m m (c c) m m (c c) m m Intersection of te lines m +c & m + c is c c m m (c c) Area (m m ) Consider now te tree lines m +c ; m +c ; m +c Also let m > m > m ABC BPQ + CPR ARQ (c c ) m m (c c) m m (c c) m m '' ' ' ( ') '' ' Tis is te required equation as it is te relation tat olds between te coordinates of an point ling on te given straigt line. EXAMPLE 8 If P(, ) and Q(5, 4) ten find te point R on -ais suc tat PR + RQ is minimum. Image of P about ais is P'(, ) Now PR P'R PR + RQ P'R + RQ will be minimum P (, ) wen R is on line P'Q. R P(, ) 4 Now of P'Q is ( 4) ( 5) 5 Q (5, 4) ais ( 0) Put 0 in it 7 7 Hence R 0, Ans. note tat te last term is negative 5. Equation of Straigt Line in Two Point form Te equation of te straigt line wic passes troug te two given points ( ', ') and ( '', '') Derivation: Te equation to te line going troug ( ', ') making an angle tan m wit OX is ' m( ')... (i) If in addition (i) passes troug te point ( '', ''), ten giving '' ' m( '' '), '' ' m. '' ' Substituting tis value in (i), we get as te required equation ANGLE BETWEEN TWO LINES Let te two straigt lines be AL and AL, meeting te ais of in L and L. Let te equations are m c and m c L L We terefore ave Y C C tan ALX m, and tan ALX m. Now LAL ALX ALX. O tan L AL tan[al X AL X] tan AL X tan AL X m m tan AL X.tan AL X m m A X. 9

24 Hence te required angle LAL m m tan m m Tus angle between two lines and m m b tan m m Note: is given Let m, m, m are te slopes of tree lines L 0;L 0;L 0 were m > m > m ten te interior angles of te ABC formed b tese lines are given b, m m tan A m m m m ; tan B m m Condition for two lines being parallel Let te straigt lines be m c, and m c. m & tan C m m m m m If be te angle between tem we ave, tan m m Two straigt lines are parallel ten te angle between tem is zero and terefore te tangent of tis angle is zero. Tus we get m m. Equation of a line parallel to a given line Te equation of a line parallel to a given line a b c 0 is a b 0, were is a constant. Proof: Let m be te slope of te line a b c 0. Ten, a m b Since te required line is parallel to te given line, te slope of te required line is also m. Let c be te -intercept of te required line. Ten its equation is m c a c b c b bc 0 a b 0, were bc constant. Note: To write a line parallel to a given line keep te epression containing and same and simpl replace te given constant b a new constant. Te value of is to be determined b some given condition. Condition for two lines being perpendicular Let te straigt lines be m c, and m c. m m If be te angle between tem we ave, tan m m If te lines be perpendicular, ten 90, and terefore tan. Te rigt and member of above equation must terefore be infinite, and tis can onl appen wen its denominator is zero. Te condition of perpendicularit is terefore tat mm 0, i.e. mm. Te straigt line m c is terefore perpendicular to m c, if m. m Equation of a line perpendicular to a given line Te equation of a line perpendicular to a given line a b c 0 is b a 0, were is a constant. Proof: Let m be te slope of te given line and m be te slope of a line perpendicular to te given line. Ten, a m and b b mm m m a. Let c be te -intercept of te required line. Ten its equation is m c b c a b a ac 0 b a 0, were ac constant. Note: To write a line perpendicular to a given line intercange te coefficients of and and cange te sign between and term and replace te given constant b a new constant. Te value of is to be determined b some given condition. EXAMPLE 9 Find te equation of te straigt line tat passes troug te point (, 4) and perpendicular to te line Te equation of a line perpendicular to is + 0 Tis passes troug te point (, 4) l 6...(i) Putting l 6 in (i), we get + 6 0, wic is te required equation. EXAMPLE 0 If AD, BE and CF are te altitudes of a triangle ABC wose verte A is te point ( 4, 5). Te coordinates of te points E and 0

25 F are (4, ) and (, 4) respectivel, ten find te equation of BC. Slope of AF 5 ( 4) 9 4 Slope of CF, Also F lies on CF Equation of CF: 0 () slope of AE Slope of BE, Also E lies on BE Equation of BE: 7 0 () Solving () and (), we get point O(, -) Since CB is passing troug intersection of AC & FC Equation of CB: ( ) 0 (can be solved witout using famil) or (l + ) + ( l) (6 + ) 0 (i) 5 4 slope of CB ; also slope of AO 4 4 since CB AO; or ; Equation of BC: Image of a point in line If te image of te point (, k ) wit respect to te line a + b + c 0 is te point (, k ) ten k k a bk c a b a b Proof: As Q (, k ) is te image of P(, ) wit respect to te line a + b + c 0 we must ave PQ perpendicular to a + b c 0 and PR RQ (see fig.) were R is te point of intersection of PQ wit te straigt line a + b + c 0. k - k Slope PQ Slope (a + b + c) i.e., - Tis implies tat k-k b - a (sa) -a b Ten + a and k k + b. Tis gives te midpoint R of PQ as + +aλ k +k +bλ, +aλ k +bλ R is te point, Now R lies on a + b + c 0 gives + aλ a + k + bλ + c 0 (a (a + b +bk +c) ) l (a + bk + c) or a +b Tus - a EXAMPLE k-k b a +bk +c a +b Find te image of te point P(, ) in te line mirror Let image of P is Q. PM MQ & PQ AB Let Q is (, k) k M is, It lies on k or k 0...(i) slope of PQ PQ AB k + k 0....(ii) k. soving (i) & (ii), we get, k Image of P(, ) is Q,

26 EXAMPLE Let A(, 0) & B, 4 be two fied points and P be a point on te line + 4, for wic location of P, AP BP is maimum? Image of A in te given line is 4 & Now for AP BP to be maimum B and image of A i.e. A 4, 9 must be collinear wit P. Hence P is point of intersection of te given line + 4 and BA i. e. 8. Required point is, LINES EQUALLY INCLINED TO GIVEN LINE Let te given straigt line is m c sown b LMN, making an angle wit te ais of suc tat tan m.in te figure Y O L P(', ') M N R Let P ( ', ') be te given point In general (ecept wen a is a rigt angle or zero) tere are two straigt lines PMR and PNS making an angle a wit te given line. Let tese lines meet te ais of in R and S and let tese make angles and wit te positive direction of te ais of. Te equations of te two required straigt lines are ' tan ( ')... (i) and ' tan ' ( ')... (ii) Now LMR RLM and ' LNS SLN (80 ). Hence tan tan tan m tan tan( ) tan tan m tan and tan ' tan(80 ) tan tan m tan tan( ) tan tan m tan. S L' X On substituting tese values in (i) and (ii), we ave as te required equations. m tan ' ( ') m tan m tan ' ( '). m tan EXAMPLE and Two equal sides of an isosceles triangle are given b te equations and + 0 and its tird side passes troug te point (, 0). find te possible equation(s) of te tird side. m AB 7 and m AC Let slope of BC (tird side) be m. Now 7 m 7m m m 7 m m ± 7m m Taking '+' sign, we get m + 0 (no real values of m possible) And taking ' ' sign, we get m + 8m 0 (m )(m + ) 0 m, Required equation of tird side can be 0 or EXAMPLE 4 In a triangle ABC, if A (, ) and and are equations of an altitude and an angle bisector respectivel drawn from B, ten find te equation of BC. BD and BE are intresect at B Coordinates of B are (, ) m AB /5 tan 5 0 m m

27 m m m or + m m /5 (rejected) or 5 equation of BC + 5 ( + ) CONCURRENCY OF THREE LINES If tree or more lines ave a common point of intersection ten te lines are called concurrent. In tis section we intend to find te necessar and sufficient condition so tat te tree lines ai bi ci 0, i,, ma be concurrent. First Metod: On solving simultaneousl We get a b c 0 and a b c 0 b c a b b c c a a b, a b c a a b. Now tis point lies on te tird line a b c 0 if and onl if bc bc ca ca a b c 0 ab ab ab ab a (b c b c ) b (c a c a ) c (a b a b ) 0 using te determinant notation introduced in sec 4.0 tis condition ma be written as a b c a b c 0 a b c EXAMPLE 5 If te lines ; and + a + b 0, were a + b, are concurrent ten find a & b. Lines are ; and + a + b 0, were a + b 4 4 a b 0 (b 4a) (4b 4) + (4a ) b 4a 4b a b + 0 b a ± EQUATION OF STRAIGHT LINE IN PARAMET- RIC FORM Derivation: Let L be te straigt line passing troug A(, ) and making an angle wit positive -ais. FromAQP we get AQ cos AP O A(, ) PQ and sin. AP P(, ) If we denote te algebraic distance AP as r Note tat for points on L on one side of A te algebraic distance is taken as positive and for points on te oter side, te algebraic distance is taken as negative. Ten we get OR AQ PQ r cos sin cos sin r Tis is te parametric equation of te line L. An point P on L is of te form ( r cos, rsin ) were r is te algebraic distance of P from A(, ). r cos, rsin is te parametric equation to L. Wen we varies r over te real numbers are gets all te points on te straigt line L. EXAMPLE 6 Find te distance between A(, ), on te line of gradient /4 and te point of intersection, P, of tis line wit A simple metod is to find te coordinates of P. Te line troug A of gradient /4 is ( ) or Q

28 Te coordinates of P are found b solving tis equation and ; te result is -, (6 + 9) and conse- 0 Ten AP + 4 quentl AP 55/4. EXAMPLE If te line cuts te curve 4 + a + b + c + d at A, B, C and D, ten find OA.OB.OC.OD ( were O is te origin). Te line cuts te given curve, ten 4 r r can be written as,. If tis line r ar br cr dr Terefore OA. OB.OC.OD r r r r 4 r r r r 4 96 EXAMPLE 8 A variable straigt line is drawn troug te point P(, ) to meet te lines +, + 4 & 4 + respectivel in points A, B & C. Find te greatest value of PB PC PA Let te slope of variable line be tan, ten an point on tis line at a distance r from P will be r cos, r sin. For r PA tis point lies on +, ence PA cos sin For r PB tis point lies on + 4, 5 ence PB cos 4sin For r PC tis point lies on 4 +, ence PA 5 4cos sin Now cos 4sin PB PC PA Hence te greatest value of 7 is 5. PB PC PA EQUATION OF STRAIGHT LINE IN DETERMINANT FORM Equation of lines written in determinant form can be ver useful sometimes because of inerent properties of determinants. For writing equation of line in determinant form we are te concept of collinearit of tree points. We know tat area of triangle formed b tree collinear points is equal to zero. Tus for an line we tr to get tree points over te line and appling te condition tat area of triangle formed b tem is equal to zero given as equation of line in determinant form. Te procedure will become clear wit following illustrations. EXAMPLE9 Find te equation of line passing troug points (, ) and (, ) in determinant form. Let (, ) is an point on te line. Now (, ) and (, ) and (, ) are tree collinear points ten area of triangle formed b tem sould be zero. 0 Tis is te required equation of line as an point, ling on te line will satisf above condition. EXAMPLE 40 Find te equation of median troug verte (, ) of a triangle in determinant form. Oter two vertices are (, ) and (, ). Let te triangle and median be as sown in te figure. Coordi- nates of D, B(, ) A(, ) D (mid point of B & C). C(, ) An point (, ) ling on te median will be collinear wit A & D. tus 0 0 tis is required equation of line. Tis is te required equation as it is te relation tat olds between te coordinates of an point ling on te given straigt line. EXAMPLE 4 Find te equation of internal and eternal angle bisectors of angle A of a triangle wit vertices A(, ), B(, ) and C(, ) in determinant form. As sown in te figure te coordinate of D i.e. te point divid- 4