STUDY PACKAGE. Subject : Mathematics Topic : The Point & Straight Lines

fo/u fopkjr Hkh# tu] ugha vkjehks dke] foifr s[k NksM+s rqjar e/;e eu dj ';kea iq#"k flag ladyi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /kez izksrk l~xq# Jh jknksm+klth egkjkt STUDY PACKAGE Subject : Mathematics Topic : The Poit & Straight Lies R Idex Theory Short Revisio Exercise (Ex to 5) 4 Assertio & Reaso 5 Que from Compt Exams 6 9 Yrs Que from IIT-JEE 7 5 Yrs Que from AIEEE Studet s Name : Class Roll No : : Address : Plot No 7, III- Floor, Near Patidar Studio, Above Bod Classes, Zoe-, MP NAGAR, Bhopal : 0 90 90 777 9, 9890 5888, WhatsApp 9009 60 559 wwwtekoclassescom wwwmathsbysuhagcom

The Poit & Straight Lie Distace Formula: The distace betwee the poits A(x,y ) ad B(x,y ) is ( x x ) + ( y ) y Solved Example # Fid the value of x, if the distace betwee the poits (x, ) ad (, ) is 5 Let P(x, ) ad Q(, ) be the give poits The PQ 5 (give) ) + ( ) 5 (x ) + 9 5 x 7 or x As ( x Self practice problems : Show that four poits (0, ), (6, 7) (, ) ad (8, ) are the vertices of a rectagle Fid the coordiates of the circumceter of the triagle whose vertices are (8, 6), (8, ) ad (, ) Also fid its circumradius As (5, ), 5 Sectio Formula : x m x + x m + ; y If P(x, y) divides the lie joiig A(x, y ) & B(x, y ) i the ratio m :, the; y + y m + m m m NOTE : (i) If is positive, the divisio is iteral, but if is egative, the divisio is exteral (ii) If P divides AB iterally i the ratio m : & Q divides AB exterally i the ratio m : the P & Q are said to be harmoic cojugate of each other wrt AB Mathematically, + ie AP, AB & AQ are i HP AB AP AQ Solved Example# Fid the coordiates of the poit which divides the lie segmet joiig the poits (6, ) ad ( 4, 5) i the ratio : (i) iterally ad (ii) exterally Let P (x, y) be the required poit (i) For iteral divisio : (ii) 4 + 6 5 + x ad y + + or x 0 ad y 5 So the coordiates of P are 0, 5 As For exteral divisio 4 6 5 x ad y or x 4 ad y 9 So the coordiates of P are ( 4, 9) As Solved Example # Fid the coordiates of poits which trisect the lie segmet joiig (, ) ad (, 4) Let A (, ) ad B(, 4) be the give poits Let the poits of trisectio be P ad Q The AP PQ QB λ (say) PB PQ + QB λ ad AQ AP + PQ λ AP : PB λ : λ : ad AQ : QB λ : λ : So P divides AB iterally i the ratio : while Q divides iterally i the ratio : + 4 + the coordiates of P are, or, 0 + + + 4 + ( ) 5 ad the coordiates of Q are, or, + + 5 Hece, the poits of trisectio are, 0 ad, As Self practice problems : I what ratio does the poit (, ) divide the lie segmet j oiig the poits (4, 4) ad 4 (7, 7)? As 5 : 8 exterally The three vertices of a parallelogram take i order are (, 0), (, ) ad (, ) respectively Fid the coordiates of the fourth vertex As (, ) Cetroid, Icetre & Excetre: If A (x, y ), B(x, y ), C(x, y ) are the vertices of triagle ABC, whose sides BC, CA, AB are of legths a, b, c respectively, the the co-ordiates of the special poits of triagle ABC are as follows : x + x + x y + y + y Cetroid G, ax+ bx+ cx ay+ by+ cy Icetre I ax + bx + cx ay + by+ cy, + c + c,ad Excetre (to A) I, + c + c ad so o NOTE : (i) Icetre divides the agle bisectors i the ratio, (b + c) : a; (c + a) : b & () : c (ii) Icetre ad excetre are harmoic cojugate of each other wrt the agle bisector o which they lie (iii) Orthoceter, Cetroid & Circumceter are always colliear & cetroid divides the lie joiig orthocetre & circumceter i the ratio : (iv) I a isosceles triagle G, O, Ι & C lie o the same lie ad i a equilateral triagle, all these four poits coicide Sol Ex 4 Fid the coordiates of (i) cetroid (ii) i-cetre of the triagle whose vertices are (0, 6), (8, ) ad (8, 0) Solutio (i) We kow that the coordiates of the cetroid of a triagle whose agular poits are (x, y ), (x, y ) x + x + x y + y + y (x, y ) are, Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page of 4

(ii) 0 + 8 + 8 6 + + 0 So the coordiates of the cetroid of a triagle whose vertices are (0, 6), (8, ) ad (8, 0) are, 6, 6 As Let A (0, 6), B (8, ) ad C(8, ) be the vertices of triagle ABC The c AB ( 0 8) + (6 ) 0, b CA ( 0 8) + (6 0) 0 or FREE Dowload Study Package from website: wwwtekoclassescom & wwwmathsbysuhagcom ad a BC ( 8 8) + ( 0) ax + bx + cx ay + by + cy The coordiates of the i-cetre are, + c + c 0 + 0 8 + 0 8 6 + 0 + 0 0 or, + 0 + 0 + 0 + 0 60 9 or, or (5, 6) As Self practice problems : 5 Two vertices of a triagle are (, 5) ad ( 7, 4) If the cetroid is (, ), fid the third vertex As(0, ) 6 Fid the coordiates of the cetre of the circle iscribed i a triagle whose vertices are ( 6, 7), (0, 7) ad (0, 8) As (, 0) 4 Area of a Triagle: If A(x, y ), B(x, y ), C(x, y ) are the vertices of triagle ABC, the its area is equal to x y ABC x y, provided the vertices are cosidered i the couter clockwise sese The above formula will give x y a ( ) ve area if the vertices (x i, y i ), i,, are placed i the clockwise sese NOTE : Area of -sided polygo formed by poits (x, y ) ; (x, y ); (x, y ) is give by x x x x x x x x + + + y y y y y y y y Solved Example # 5: If the coordiates of two poits A ad B are (, 4) ad (5, ) respectively Fid the coordiates of ay poit P if PA PB ad Area of PAB 0 Solutio Let the coordiates of P be (x, y) The PA PB PA PB (x ) + (y 4) (x 5) + (y + ) x y 0 x y 4 Now, Area of PAB 0 5 ± 0 6x + y 6 ± 0 6x + y 46 0 x + y 0 or or 6x + y 6 0 x + y 0 Solvig x y 0 ad x + y 0 we get x 7, y Solvig x y 0 ad x + y 0, we get x, y 0 Thus, the coordiates of P are (7, ) or (, 0) As Self practice problems : 7 The area of a triagle is 5 Two of its vertices are (, ) ad (, ) The third vertex lies o 7 y x + Fid the third vertex As, or, 8 The vertices of a quadrilateral are (6, ), (, 5), (4, ) ad (x, x) ad are deoted by A, B, C ad D, respectively Fid the values of x so that the area of triagle ABC is double the area of triagle DBC As x or 8 8 5 Slope Formula: If θ is the agle at which a straight lie is iclied to the positive directio of x axis, & 0 θ < 80, θ 90, the the slope of the lie, deoted by m, is defied by m ta θ If θ is 90, m does ot exist, but the lie is parallel to the y axis If θ 0, the m 0 & the lie is parallel to the x-axis If A (x, y ) & B (x, y ), x x, are poits o a straight lie, the the slope m of the lie is give by : y m y x x Solved Example # 6: (i) 0º What is the slope of a lie whose icliatio is : (ii) 90º (iii) 0º (iv) 50º Solutio (i) Here θ 0º Slope ta θ ta 0º 0 As (ii) Here θ 90º The slope of lie is ot defied As (iii) Here θ 0º Slope ta θ ta 0º ta (80º 60º) ta 0º As (iv) Here θ 50º Slope ta θ ta 50º ta (80º 0º) ta 0º As Solved Example # 7 : Fid the slope of the lie passig through the poits : (i) (, 6) ad ( 4, ) (ii) (5, 9) ad (, 9) Solutio (i) Let A (, 6) ad B ( 4, ) Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page of 4 6 4 4 y Slope of AB As 4 5 5 y Usig slope x x (ii) Let A (5, 9), B (, 9) 9 9 0 Slope of AB 5 0 As

Self practice problems : 9 Fid the value of x, if the slope of the lie joiig (, 5) ad (x, 7) is 4 As 0 What is the icliatio of a lie whose slope is (i) 0 (ii) (iii) (iv) / As (i) 0º, (ii) 45º, (iii) 5º, (iv) 50º 6 Coditio of colliearity of three poits: Poits A (x, y ), B (x, y ), C(x, y ) are colliear if x y y (i) m AB m BC m CA ie y y x x y x (ii) ABC 0 ie y 0 x x x y (iii) AC AB + BC or AB ~ BC (iv) A divides the lie segmet BC i some ratio Solved Example # 8 Show that the poits (, ), (, ) ad (, 5) are colliear Let (, ) (, ) ad (, 5) be the coordiates of the poits A, B ad C respectively 5 Slope of AB ad Slope of BC Slope of AB slope of AC AB & BC are parallel A, B, C are colliear because B is o both lies AB ad BC Prove that the poits (a, 0), (0, b) ad (, ) are colliear if a + b 7 Equatio of a Straight Lie i various forms: (i) Poit - Slope form : y y m (x x ) is the equatio of a straight lie whose slope is m & which passes through the poit (x, y ) Solved Example # 9 : Fid the equatio of a lie passig through (, ) ad iclied at a agle of 5º with the positive directio of x-axis Here, m slope of the lie ta 5º ta (90º + 45º) cot 45º, x, y So, the equatio of the lie is y y m (x x ) ie y ( ) (x ) or y + x + or x + y + 0 As Fid the equatio of the perpedicular bisector of the lie segmet j oiig the poits A(, ) ad B (6, 5) As x y 6 0 (ii) Slope itercept form : y mx + c is the equatio of a straight lie whose slope is m & which makes a itercept c o the y axis Solved Example # 0: Fid the equatio of a lie with slope ad cuttig off a itercept of 4 uits o egative directio of y-axis Here m ad c 4 So, the equatio of the lie is y mx + c ie y x 4 or x + y + 4 0 As Fid the equatio of a straight lie which cuts off a itercept of legth o y-axis ad is parallel to the lie joiig the poits (, ) ad (, 4) As x + y 0 y y (iii) Two poit form : y y x x (x x ) is the equatio of a straight lie which passes through the poits (x, y ) & (x, y ) Solved Example # Fid the equatio of the lie joiig the poits (, ) ad (4, ) Here the two poits are (x, y ) (, ) ad (x, y ) (4, ) So, the equatio of the lie i two-poit form is ( ) y (x + ) y x x + y 0 As 4 4 Fid the equatios of the sides of the triagle whose vertices are (, 8), (4, ) ad ( 5, ) Also fid the equatio of the media through (, 8) As x + y 6 0, x 9y 0, x 4y + 4 0, x + y + 0 x y (iv) Determiat form : Equatio of lie passig through (x, y ) ad (x, y ) is x y 0 Solved Example # Fid the equatio of lie passig through (, 4) & (, ) x y x 4 0 x y + 0 0 As 5 Fid the equatio of the passig through (, ) & (, ) As 4x + y + 5 0 (v) x y Itercept form : + is the equatio of a straight lie which makes itercepts a & b o OX & OY a b respectively Solved Example # : Fid the equatio of the lie which passes through the poit (, 4) ad the sum of its itercepts o the axes is 4 Sol Let the equatio of the lie be a x + b y (i) y Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 4 of 4 This passes through (, 4), therefore a + b 4 (ii) 4 It is give that 4 b 4 a Puttig b 4 a i (ii), we get + a 4 a a a + 4 0 (a 7) (a 6) 0 a 7, 6 For a 7, b 4 7 7 ad for a 6, b 4 6 8 Puttig the values of a ad b i (i), we get the equatios of the lies x y y y + ad + or x + y 7 ad 4x + y 4 As 7 7 6 8

6 Fid the equatio of the lie through (, ) so that the segmet of the lie itercepted betwee the axes is bisected at this poit As x + y (vi) Perpedicular/Normal form : xcos α + ysi α p (where p > 0, 0 α < π) is the equatio of the straight lie where the legth of the perpedicular from the origi O o the lie is p ad this perpedicular makes a agle α with positive x axis Solved Example # 4: Fid the equatio of the lie which is at a distace from the origi ad the perpedicular from the origi to the lie makes a agle of 0º with the positive directio of the x-axis Here p, α 0º Equatio of the lie i the ormal form is y x cos 0º + y si 0º or x + or x + y 6 As 7 The legth of the perpedicular from the origi to a lie is 7 ad the lie makes a agle of 50º with the positive directio of y-axis Fid the equatio of the lie As x + y 4 0 x x y y (vii) Parametric form : P (r) (x, y) (x + r cos θ, y + r si θ) or cosθ siθ r is the equatio of the lie i parametric form, where r is the parameter whose absolute value is the distace of ay poit (x, y) o the lie from the fixed poit (x, y ) o the lie Solved Example # 5: Fid the equatio of the lie through the poit A(, ) ad makig a agle of 45º with the x-axis Also determie the legth of itercept o it betwee A ad the lie x + y + 0 The equatio of a lie through A ad makig a agle of 45º with the x-axis is x y x y or or x y + 0 cos 45º si45º Suppose this lie meets the lie x + y + 0 at P such that AP r The the coordiates of P are give by x y r x + r cos 45º, y + r si 45º cos 45º si45º r r x +, y + r r Thus, the coordiates of P are +, + r r Sice P lies o x + y + 0, so + + + + 0 r 6 r legth AP r Thus, the legth of the itercept As 8 A straight lie is draw through the poit A (, ) the straight lie (viii) makig a agle of π/6 with positive directio of the x-axis If it meets x 4y + 8 0 i B, fid the distace betwee A ad B As 6 uits Geeral Form : ax + by + c 0 is the equatio of a straight lie i the geeral form I this case, slope of lie b a x - itercept a c y - itercept b c Solved Example # 6 Fid slope, x-itercept & y-itercept of the lie x y + 5 0 Here, a, b, c 5 a slope As b x-itercept a c 5 5 y-itercept As As 9 Fid the slope, x-itercept & y-itercept of the lie x 5y 8 0 As 8 8,, 5 5 8 Agle betwee two straight lies i terms of their slopes: If m & m are the slopes of two itersectig straight lies (m m ) & θ is the acute agle betwee them, the ta θ m m + mm NOTE : (i) Let m, m, m are the slopes of three lies L 0;L 0;L 0 where m > m > m the the iterior agles of the ABC foud by these lies are give by, m m m m m ta A ; ta B & ta C m + mm + mm + mm (ii) The equatio of lies passig through poit (x, y ) ad makig agle α with the lie y mx + c are give by : (y y ) ta (θ α) (x x ) & (y y ) ta (θ + α) (x x), where ta θ m Solved Example # 7: The acute agle betwee two lies is π/4 ad slope of oe of them is / Fid the slope of the other lie If θ be the acute agle betwee the lies with slopes m ad m, the ta θ m m + m m Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 5 of 4

Let θ 4 π ad m m π m ta 4 + m + m m m Now + m m ad + m m + m + or m The slope of the other lie is either / or As Solved Example # 8: Fid the equatio of the straight lie which passes through the origi ad makig agle 60º with the lie x + y + 0 Give lie is x + y + 0 y x Slope of () Let slope of the required lie be m Also betwee these lies is give to be 60º ta 60º m + m ( / ) ( / ) m + m m + ± m m + m + m m m Usig y mx + c, the equatio of the required lie is y x + 0 ie x y 0 ( This passes through origi, so c 0) m + m + + m m m is ot defied The slope of the required lie is ot defied Thus, the required lie is a vertical lie This lie is to pass through the origi The equatio of the required lie is x 0 As 0 A vertex of a equilateral triagle is (, ) ad the equatio of the opposite side is x + y Fid the equatio of the other sides of the triagle As ( + )x y + 0 ad ( + ) x y 0 9 Parallel Lies: (i) Whe two straight lies are parallel their slopes are equal Thus ay lie parallel to y mx + c is of the type y mx + d, where k is a parameter a (ii) Two lies ax + by + c 0 ad a x + b y + c 0 are parallel if a b b c c Thus ay lie parallel to ax + by + c 0 is of the type ax + by + k 0, where k is a parameter (iii) The distace betwee two parallel lies with equatios ax + by + c 0 & c c ax + by + c 0 is NOTE: Coefficiets of x & y i both the equatios must be same p p (iv) The area of the parallelogram siθ, where p & p are distaces betwee two pairs of opposite sides & θ is the agle betwee ay two adjacet sides Note that area of the parallelogram bouded by the lies y m x + c, y (c c )(d d ) m x + c ad y m x + d, y m x + d is give by m m Sol Ex 9: x y 7 Give lie is x y 7 Fid the equatio of the straight lie that has y-itercept 4 ad is parallel to the straight lie () y x 7 y x 7 Slope of () is / The required lie is parallel to (), so its slope is also /, y-itercept of required lie 4 By usig y mx + c form, the equatio of the required lie is y x + 4 or x y + 0As Solved Example # 0: Two sides of a square lie o the lies x + y ad x + y + 0 What is its area? Clearly the legth of the side of the square is equal to the distace betwee the parallel lies x + y 0 (i) ad x + y + 0 (ii) Puttig x 0 i (i), we get y So (0, ) is a poit o lie (i) Now, Distace betwee the parallel lies 0 + + legth of the from (0, ) to x + y + 0 + Thus, the legth of the side of the square is ad hece its area 9 Solved Example # : Fid the area of the parallelogram whose sides are x + y + 0, x + 4y 5 0, x + 4y + 5 0 ad x + 4y 0 0 Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 6 of 4

5 0 5 Here, c, c, d, d, m, m 4 5 0 5 + + 70 Area sq uits As + 4 Fid the area of parallelogram whose sides are give by 4x 5y + 0, x y 6 0, 5 4x 5y 0 ad x 6y + 5 0 As sq uits 4 0 Perpedicular Lies: (i) W he two lies of slopes m & m are at right agles, the product of their slopes is, ie m m Thus ay lie perpedicular to y mx + c is of the form y m x + d, where d is ay parameter (ii) Two lies ax + by + c 0 ad a x + b y + c 0 are perpedicular if aa + bb 0 Thus ay lie perpedicular to ax + by + c 0 is of the form bx ay + k 0, where k is ay parameter Solved Example # Fid the equatio of the straight lie that passes through the poit (, 4) ad perpedicular to the lie x + y + 5 0 The equatio of a lie perpedicular to x + y + 5 0 is x y + λ 0 (i) This passes through the poit (, 4) 4 + λ 0 λ 6 Puttig λ 6 i (i), we get x y + 6 0, which is the required equatio As Aliter The slope of the give lie is / Sice the required lie is perpedicular to the give lie So, the slope of the required lie is / As it passes through (, 4) So, its equatio is y 4 (x ) or x y + 6 0 As The vertices of a triagle are A(0, 4), B ( 4, 9) ad C(, ) Fid the equatio of its altitudes Also fid its orthocetre 9 As x 5y + 0 0, x + 5y + 0, 4x 5y + 0,, 5 Positio of the poit (x, y ) relative of the lie ax + by + c 0: If ax + by + c is of the same sig as c, the the poit (x, y ) lie o the origi side of ax + by + c 0 But if the sig of ax + by + c is opposite to that of c, the poit (x, y ) will lie o the o origi side of ax + by + c 0 I geeral two poits (x, y ) ad (x, y ) will lie o same side or opposite side of ax + by + c 0 accordig as ax + by + c ad ax + by + c are of same or opposite sig respectively Solved Example # Show that (, 4) ad (0, ) lie o the opposite sides of the lie x + y + 7 0 At (, 4), the value of x + y + 7 + (4) + 7 0 > 0 At (0, ), the value of x + y + 7 0 + ( ) + 7 < 0 The poits (, 4) ad (0, ) are o the opposite sides of the give lie As Self practice problems : Are the poits (, 4) ad (, 6) o the same or opposite side of the lie x 4y 8? As Opposite sides 4 Which oe of the poits (, ), (, ) ad (, ) lies o the side of the lie 4x + y 5 0 o which the origi lies? As (, ) The ratio i which a give lie divides the lie segmet joiig two poits: m ax+ by+ c Let the give lie ax + by + c 0 divide the lie segmet joiig A(x, y ) & B(x, y ) i the ratio m :, the ax+ by+ c If A & B are o the same side of the give lie the m/ is egative but if A & B are o opposite sides of the give lie, the m/ is positive Solved Example # 4 Fid the ratio i w hich the lie joiig the poits A (, ) ad B(, 4) is divided by the lie x + y 5 0 Let the lie x + y 5 divides AB i the ratio k : at P coordiate of P are k + 4k +, k + k + Sice P lies o x + y 5 0 k + 4k + + 5 0 k k + k + Required ratio is : extreally As Aliter Let the ratio is m : m ( + 5) ratio is : exterallyas ( ) + 4 5 5 If the lie x y + λ 0 divides the lie joiig the poits A (, ) & B(, ) iterally i the ratio :, fid λ 8 As 5 Legth of perpedicular from a poit o a lie: The legth of perpedicular from P(x, y ) o ax + by + c 0 is a x + by + c Solved Example # 5 Fid the distace betwee the lie x 5y + 9 0 ad the poit (, ) 5 + 9 4 5 + 9 8 The required distace As Successful People Replace the words like; + "wish", ( 5) "try" & "should" with "I Will" Ieffective People do't Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 7 of 4

Solved Example # 6 Fid all poits o x + y 4 that lie at a uit distace from the lie 4x + y 0 0 Note that the coordiates of a arbitrary poit o x + y 4 ca be obtaied by puttig x t (or y t) ad the obtaiig y (or x) from the equatio of the lie, where t is a parameter Puttig x t i the equatio x + y 4 of the give lie, we obtai y 4 t So, coordiates of a arbitrary poit o the give lie are P(t, 4 t) Let P(t, 4 t) be the required poit The, distace of P from the lie 4x + y 0 0 is uity ie 4t + (4 t) 0 4 + t + 5 t + ± 5 t 7 or t Hece, required poits are ( 7, ) ad (, ) As 6 Fid the legth of the altitudes from the vertices of the triagle with vertices :(, ), (5, ) ad (, ) 6 8 6 As,, 5 7 4 Reflectio of a poit about a lie: x x y y ax+ by+ c (i) Foot of the perpedicular from a poit o the lie is a b (ii) The image of a poit (x, y ) about the lie ax + by + c 0 is x x y y ax+ by+ c a b Solved Example # 7 Fid the foot of perpedicular of the lie draw from P (, 5) o the lie x y + 0 Slope of PM Equatio of PM is x + y 0 (i) solvig equatio (i) with x y + 0, we get coordiates of M (0, ) As x + y 5 ( ( ) + ( ) 5 + ) Aliter Here, () + ( ) x + y 5 x + x 0 ad y 5 y M is (0, ) As Solved Example # 8 Fid the image of the poit P(, ) i the lie mirror x y + 4 0 Let image of P is Q Aliter Let PM MQ & PQ AB Q is (h, k) h k + M is, It lies o x y + 4 0 h k + + 4 0 or h k 0 (i) k slope of PQ h + PQ AB k h + h + k 0 (ii) sovig (i) & (ii), we get h, k Image of P(, ) is Q, As The image of P (, ) about the lie x y + 4 0 is x + y [( ) () + 4] + ( ) x + y 8 x + 6 x & y 6 4 y image is, As Self practice problems : 4 7 Fid the foot of perpedicular of the lie draw from (, ) o the lie x y 0 As, 8 Fid the image of the poit (, ) i y-axis As (, ) Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 8 of 4

5 Bisectors of the agles betwee two lies: Equatios of the bisectors of agles betwee the lies ax + by + c 0 & ax + by + c a x + b y + c a x + b y + c 0 (ab a b) are : ± a + b a + b NOTE : Equatio of straight lies passig through P(x, y ) & equally iclied with the lies a x + b y + c 0 & a x + b y + c 0 are those which are parallel to the bisectors betwee these two lies & passig through the poit P Solved Example # 9 Fid the equatios of the bisectors of the agle betwee the straight lies x 4y + 7 0 ad x 5y 8 0 The equatios of the bisectors of the agles betwee x 4y + 7 0 ad x 5y 8 0 are x 5y 8 + ( 5) x 4y + 7 (ii) Sice 9x 7y 4 is the bisector of the obtuse agle betwee the give lies, therefore the other bisector 7x + 9y 0 will be the bisector of the acute agle betwee the give lies d Method : Writig the equatio of the lies so that costats become positive we have 4x y + 6 0 () + ( 4) x 4y + 7 x 5y 8 or ± 5 or 9x 5y + 9 ± (60 x 5 y 8) Takig the positive sig, we get x + 7 y 0 as oe bisector As Takig the egative sig, we get 99 x 77 y + 5 0 as the other bisector As 9 Fid the equatios of the bisectors of the agles betwee the followig pairs of straight lies x + 4y + 0 ad x 5y + 0 As x 77y 9 0 ad 99x + 7y + 9 0 6 Methods to discrimiate betwee the acute agle bisector & the obtuse agle bisector: (i) If θ be the agle betwee oe of the lies & oe of the bisectors, fid ta θ If ta θ <, the θ < 90 so that this bisector is the acute agle bisector If ta θ >, the we get the bisector to be the obtuse agle bisector (ii) Let L 0 & L 0 are the give lies & u 0 ad u 0 are the bisectors betwee L 0 & L 0 Take a poit P o ay oe of the lies L 0 or L 0 ad drop perpedicular o u 0 & u 0 as show If, p < q u is the acute agle bisector p > q u is the obtuse agle bisector p q the lies L & L are perpedicular (iii) If aa + bb < 0, the the equatio of the bisector of this acute agle is a x + by + c a + x + b y + c a + b If, however, aa + bb > 0, the equatio of the bisector of the obtuse agle is : a x + by + c a x + b y + c + a + b Solved Example # 0 For the straight lies 4x + y 6 0 ad 5x + y + 9 0, fid the equatio of the (i) bisector of the obtuse agle betwee them; (ii) bisector of the acute agle betwee them; (i) The equatios of the give straight lies are 4x + y 6 0 () 5x + y + 9 0 () The equatio of the bisectors of the agles betwee lies () ad () are 4x + y 6 5x + y + 9 4x + y 6 5 x + y + 9 ± or ± 4 + 5 + 5 Takig the positive sig, we have 4x + y 6 5 x + y + 9 5 or 5x + 9y 78 5x + 60y + 45 or 7x y 0 or 9x 7y 4 0 Takig the egative sig, we have 4x + y 6 5 x + y + 9 5 or 5x + 9y 78 5x 60y 45 or 77x + 99y 0 or 7x + 9y 0 Hece the equatio of the bisectors are 9x 7y 4 0 () ad 7x + 9y 0 (4) Now slope of lie () 4 ad slope of the bisector () 7 9 If θ be the acute agle betwee the lie () ad the bisector (), the ta θ 9 4 + 7 9 4 + 7 7 + 8 6 55 5 > θ > 45º Hece 9x 7y 4 0 is the bisector of the obtuse agle betwee the give lies () ad () As ± Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 9 of 4

ad 5x + y + 9 0 () Here a 4, a 5, b, b Now a a + b b 0 6 56 < 0 origi does ot lie i the obtuse agle betwee lies () ad () ad hece equatio of the bisector of the obtuse agle betwee lies () ad () will be 4x y + 6 5x + y + 9 ( 4) + ( ) 5 + or ( 4x y + 6) 5(5x + y + 9) or 7x y 0 or 9x 7y 4 0 As ad the equatio of the bisector of the acute agle will be (origi lies i the acute agle) 4x y + 6 5x + y + 9 ( 4) + ( ) 5 + or 77x + 99y 0 or 7x + 9y 0 As 0 Fid the equatios of the bisectors of the agles betwee the lies x + y 0 ad 7x y + 5 0 ad state which of them bisects the acute agle betwee the lies As x y + 0 0 (bisector of the obtuse agle); 4x + 0 (bisector of the acute agle) 7 To discrimiate betwee the bisector of the agle cotaiig a poit: To discrimiate betwee the bisector of the agle cotaiig the origi & that of the agle ot cotaiig the origi Rewrite the equatios, ax + by + c 0 & a x + b y + c 0 such that the costat terms c, c are positive The ; a x + by + c a + x + b y + c gives the equatio of the bisector of the agle cotaiig the origi & a x + by + c a + b a x + b y + c gives the equatio of the bisector of the agle ot cotaiig the origi I geeral equatio of the a + b bisector which cotais the poit (α, β) is, a x + b y + c a a + b x + b y + c or a x + b y + c a a + b a + b x + b y + c accordig as a + b a α + b β + c ad a α + b β + c havig same sig or otherwise Solved Example # For the straight lies 4x + y 6 0 ad 5x + y + 9 0, fid the equatio of the bisector of the agle which cotais the origi For poit O(0, 0), 4x + y 6 6 < 0 ad 5x + y + 9 9 > 0 Hece for poit O(0, 0) 4x + y 6 ad 5x + y + 9 are of opposite sigs Hece equatio of the bisector of the agle betwee the give lies cotaiig the origi will be 4x + y 6 5x + y + 9 (4) + () 5 + 4x + y 6 5 x + y + 9 or 5 or 5x + 9y 78 5x 60y 45 or 77x + 99y 0 or 7x + 9y 0 As Fid the equatio of the bisector of the agle betwee the lies x + y 0 ad x 6y 5 0 which cotais the poit (, ) As x 9 0 8 Coditio of Cocurrecy: Three lies a x + b y + c 0, a x + b y + c 0 & a x + b y + c 0 are cocurret if a b c a b c 0 a b c Alteratively : If three costats A, B & C (ot all zero) ca be foud such that A(a x + b y + c ) + B(a x + b y + c ) + C(a x + b y + c ) 0, the the three straight lies are cocurret Solved Example # Prove that the straight lies 4x + 7y 9, 5x 8y + 5 0 ad 9x y + 6 0 are cocurret Give lies are 4x + 7y 9 0 () ad 5x 8y + 5 0 9x y + 6 0 () () 4 7 9 5 8 5 4( 48 + 5) 7 (0 5) 9 ( 5 + 7) + 75 60 0 9 6 Hece lies (), () ad () are cocurret Proved Fid the value of m so that the lies x + y + 0, x y + 0 ad x + my 0 may be cocurret 9 As 4 Family Of Straight Lies: The equatio of a family of straight lies passig through the poit of itersectio of the lies, L a x + b y + c 0 & L a x + b y + c 0 is give by L + k L 0 ie (a x + b y + c ) + k(a x + b y + c ) 0, where k is a arbitrary real umber Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 0 of 4

NOTE : (i) If u ax + by + c, u a x + b y + d, u ax + by + c, u 4 a x + b y + d the u 0;u 0; u 0 ; u 4 0 form a parallelogram The diagoal BD ca be give by u u u u 4 0 (ii) The diagoal AC is also give by u + λu 4 0 ad u + µu 0, if the two equatios are idetical for some real λ ad µ [For gettig the values of λ & µ compare the coefficiets of x, y & the costat terms] Solved Example # Fid the equatio of the straight lie which passes through the poit (, ) ad the poit of itersectio of the lies x + y + 4 0 ad x y 8 0 Ay lie through the itersectio of the lies x + y + 4 0 ad x y 8 0 has the equatio (x + y + 4) + λ (x y 8) 0 (i) This will pass through (, ) if ( + 4) + λ (6 + 8) 0 or + λ 0 λ Puttig the value of λ i (i), the required lie is (x + y + 4) + ( ) (x y 8) 0 or 8x + 4y + 8 0 or x y 7 0 As Aliter Solvig the equatios x + y + 4 0 ad x y 8 0 by cross-multiplicatio, we get x, y 5 So the two lies itersect at the poit (, 5) Hece the required lie passes through (, ) ad (, 5) ad so its equatio is 5 + y + (x ) or x y 7 0 As Solved Example # 4 Obtai the equatios of the lies passig through the itersectio of lies 4x y 0 ad x 5y + 0 ad equally iclied to the axes The equatio of ay lie through the itersectio of the give lies is (4x y ) + λ (x 5y + ) 0 or x ( λ + 4) y (5λ + ) + λ 0 (i) λ + 4 Let m be the slope of this lie The m 5λ + As the lie is equally iclied with the axes, therefore λ + 4 m ta 45º of m ta 5º m ±, 5λ + 0 ad 4x 4y 0 ± λ or, puttig the values of λ i (i), we get x + y 4 ie x + y 0 ad x y as the equatios of the required lies As Fid the equatio of the lies through the poit of itersectio of the lies x y + 0 ad x + 5y 9 0 ad whose distace from the origi is 5 As x + y 5 0 0 A Pair of straight lies through origi: (i) A homogeeous equatio of degree two, "ax² + hxy + by² 0" always represets a pair of straight lies passig through the origi if : (a) h² > ab lies are real & distict (b) h² ab lies are coicidet (c) h² < ab lies are imagiary with real poit of itersectio ie (0, 0) (ii) (iii) If y m x & y m x be the two equatios represeted by ax² + hxy + by² 0, the; m + m h & m m a b b If θ is the acute agle betwee the pair of straight lies represeted by, ax² + hxy + by² 0,the ; ta θ h ab (iv) The coditio that these lies are : (a) At right agles to each other is 0 ie co efficiet of x² + co efficiet of y² 0 (b) Coicidet is h² ab (c) Equally iclied to the axis of x is h 0ie coeff of xy 0 NOTE : A homogeeous equatio of degree represets straight lies passig through origi (v) The equatio to the pair of straight lies bisectig the agle betwee the straight lies, ax² + hxy + by² 0 is x y xy a b h Solved Example # 5 Show that the equatio 6x 5xy + y 0 represets a pair of distict straight lies, each passig through the origi Fid the separate equatios of these lies The give equatio is a homogeeous equatio of secod degree So, it represets a pair of straight lies passig through the origi Comparig the give equatio with ax + hxy + by 0, we obtai a 6, b ad h 5 5 h ab 6 > 0 h > ab 4 4 Hece, the give equatio represets a pair of distict lies passig through the origi Now, 6x 5xy + y 0 y x y x y 5 + 6 0 x y y y + 6 0 x x x y 0 x Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page of 4 y y 0 or 0 y x 0 or y x 0 x x So the give equatio represets the straight lies y x 0 ad y x 0 As

Solved Example # 6 Fid the equatios to the pair of lies through the origi which are perpedicular to the lies represeted by x 7xy + y 0 We have x 7xy + y 0 x 6xy xy + y 0 x(x y) y (x y) 0 (x y) (x y) 0 x y 0 or x y 0 Thus the give equatio represets the lies x y 0 ad x y 0 The equatios of the lies passig through the origi ad perpedicular to the give lies are y 0 (x 0) FREE Dowload Study Package from website: wwwtekoclassescom & wwwmathsbysuhagcom ad y 0 (x 0) [ (Slope of x y 0) is / ad (Slope of x y 0) is ] y + x 0 ad y + x 0 As Solved Example # 7 Fid the agle betwee the pair of straight lies 4x + 4xy + y 0 Give equatio is 4x + 4xy + y 0 Here a coeff of x 4, b coeff of y ad h coeff of xy 4 h Now ta θ h ab 44 44 4 + Where θ is the acute agle betwee the lies 4 acute agle betwee the lies is ta ad obtuse agle betwee them is 4 4 π ta As Solved Example # 8 Fid the equatio of the bisectors of the agle betwee the lies represeted by x 5xy + y 0 Give equatio is x 5xy + y 0 () comparig it with the equatio ax + hxy + by 0 () we have a, h 5; ad b 4 Now the equatio of the bisectors of the agle betwee the pair of lies () is x y xy x y or ; or 4 5 or 5x xy 5y 0 As xy 5 x y a b Self practice problems : 7 4 Fid the area of the triagle formed by the lies y 9xy + 8x 0 ad y 9 As sq uits 4 5 If the pairs of straight lies x pxy y 0 ad x qxy y 0 be such that each pair bisects the agle betwee the other pair, prove that pq Geeral equatio of secod degree represetig a pair of Straight lies: (i) ax² + hxy + by² + gx + fy + c 0 represets a pair of straight lies if : a h g h abc + fgh af² bg² ch² 0, ie if b f 0 g f c (ii) The agle θ betwee the two lies represetig by a geeral equatio is the same as that betwee the two lies represeted by its homogeeous part oly Solved Example # 9 Prove that the equatio x + 5xy + y + 6x + 7y + 4 0 represets a pair of straight lies Fid the co-ordiates of their poit of itersectio ad also the agle betwee them Give equatio is x + 5xy + y + 6x + 7y + 4 0 Writig the equatio () as a quadratic equatio i x we have x + (5y + 6) x + y + 7y + 4 0 x (5y + 6) ± (5y + 6) ± (5y + 6) 4(y 4 4 + 7y + 4) 5y + 60y + 6 4y 56y (5y + 6) ± y + 4y + 4 ( 5y + 6) ± (y + ) 4 4 5 y 6 + y + 5y 6 y x, 4 4 or 4x + 4y + 4 0 ad 4x + 6y + 8 0 or x + y + 0 ad x + y + 4 0 Hece equatio () represets a pair of straight lies whose equatio are x + y + 0 () ad x + y + 4 0 () As Solvig these two equatios, the required poit of itersectio is (, ) As 6 Fid the combied equatio of the straight lies passig through the poit (, ) ad parallel to the lies represeted by the equatio x 5xy + 4y + x + y 0 ad fid the agle betwee them As x 5xy + 4y + x y 0, ta 5 Homogeizatio : The equatio of a pair of straight lies joiig origi to the poits of itersectio of the lie L lx + my + 0 ad a secod degree curve, S ax² + hxy + by² + gx + fy + c 0 h xy Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page of 4

lx + my lx + my lx + my is ax² + hxy + by² + gx fy + c + 0 The equatio is obtaied by homogeizig the equatio of curve with the help of equatio of lie FREE Dowload Study Package from website: wwwtekoclassescom & wwwmathsbysuhagcom NOTE : Equatio of ay curve passig through the poits of itersectio of two curves C 0 ad C 0 is give by λ C + µ C 0 where λ & µ are parameters Solved Example # 40 Prove that the agle betwee the lies joiig the origi to the poits of itersectio of the straight lie y x + with the curve x + xy + y + 4x + 8y 0 is ta Equatio of the give curve is x + xy + y + 4x + 8y 0 y x ad equatio of the give straight lie is y x ; Makig equatio () homogeeous equatio of the secod degree i x ay y with the help of (), we have y x y x y x x + xy + y + 4x + 8y 0 or x + xy + y + (4xy + 8y x 4 xy) 4 (y 6xy + 9x ) 0 or 4x + 8xy + y + (8y x 0xy) (y 6xy + 9x ) 0 or 9x + 4xy + 7y 0 or 9x 4xy 7y 0 or 7x xy y 0 This is the equatio of the lies joiig the origi to the poits of itersectio of () ad () Comparig equatio () with the equatio ax + hxy + by 0 we have a 7, b ad h ie h If θ be the acute agle betwee pair of lies (), the ta θ h ab Self practice problems : + 7 7 8 6 θ ta Proved 7 Fid the equatio of the straight lies joiig the origi to the poits of itersectio of the lie x + 4y 5 0 ad the curve x + y 5 As x y 4xy 0 8 Fid the equatio of the straight lies joiig the origi to the poits of itersectio of the lie lx + my + 0 ad the curve y 4ax Also, fid the coditio of their perpedicularity As 4alx + 4amxy + y 0; 4al + 0 Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page of 4

SHORT REVISION DISTANCE FORMULA : The distace betwee the poits A(x,y ) ad B(x,y ) is ( x x) + (y y) SECTION FORMULA : If P(x, y) divides the lie joiig A(x, y ) & B(x, y ) i the ratio m :, the ; mx + x my + y x ; y m+ m+ If m is positive, the divisio is iteral, but if m is egative, the divisio is exteral Note : If P divides AB iterally i the ratio m : & Q divides AB exterally i the ratio m : the P & Q are said to be harmoic cojugate of each other wrt AB Mathematically ; + ie AP, AB & AQ are i HP AB AP AQ CENTROID AND INCENTRE : If A(x, y ), B(x, y ), C(x, y ) are the vertices of triagle ABC, whose sides BC, CA, AB are of legths a, b, c respectively, the the coordiates of the cetroid are : x + x+ x y+ y+ y, & the coordiates of the icetre are : ax + bx+ cx ay+ by+ cy, a+ b+ c a+ b+ c Note that icetre divides the agle bisectors i the ratio (b + c) : a ; (c + a) : b & () : c REMEMBER : (i) Orthocetre, Cetroid & circumcetre are always colliear & cetroid divides the lie joiig orthocetre & cercumcetre i the ratio : (ii) I a isosceles triagle G, O, I & C lie o the same lie 4 SLOPE FORMULA : If θ is the agle at which a straight lie is iclied to the positive directio of x axis, & 0 θ < 80, θ 90, the the slope of the lie, deoted by m, is defied by m ta θ If θ is 90, m does ot exist, but the lie is parallel to the y axis If θ 0, the m 0 & the lie is parallel to the x axis If A (x, y ) & B (x, y ), x x, are poits o a straight lie, the the slope m of the lie is give by: y m y x x 5 CONDITION OF COLLINEARITY OF THREE POINTS (SLOPE FORM) : y y y y Poits A (x, y ), B (x, y ), C(x, y ) are colliear if x x x x 6 EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS : (i) Slope itercept form: y mx + c is the equatio of a straight lie whose slope is m & which makes a itercept c o the y axis (ii) Slope oe poit form: y y m (x x ) is the equatio of a straight lie whose slope is m & which passes through the poit (x, y ) (iii) Parametric form : The equatio of the lie i parametric form is give by x x y y cosθ siθ r (say) Where r is the distace of ay poit (x, y) o the lie from the fixed poit (x, y ) o the lie r is positive if the poit (x, y) is o the right of (x, y ) ad egative if (x, y) lies o the left of (x, y ) (iv) Two poit form : y y y y (x x x x ) is the equatio of a straight lie which passes through the poits (x, y ) & (x, y ) x y (v) Itercept form : + is the equatio of a straight lie which makes itercepts a & b a b o OX & OY respectively (vi) Perpedicular form : xcos α + ysi α p is the equatio of the straight lie where the legth of the perpedicular from the origi O o the lie is p ad this perpedicular makes agle α with positive side of x axis (vii) Geeral Form : ax + by + c 0 is the equatio of a straight lie i the geeral form 7 POSITION OF THE POINT (x, y ) RELATIVE TO THE LINE ax + by + c 0 : If ax + by + c is of the same sig as c, the the poit (x, y ) lie o the origi side of ax + by + c 0 But if the sig of ax + by + c is opposite to that of c, the poit (x, y ) will lie o the o-origi side of ax + by + c 0 8 THE RATIO IN WHICH A GIVEN LINE DIVIDES THE LINE SEGMENT JOINING TWO POINTS : Let the give lie ax + by + c 0 divide the lie segmet joiig A(x, y ) & B(x, y ) i the ratio m :, the m a x + by + c If A & B are o the same side of the give lie the m is egative but if A & B a x + by + c are o opposite sides of the give lie, the m is positive 9 LENGTH OF PERPENDICULAR FROM A POINT ON A LINE : The legth of perpedicular from P(x, y ) o ax + by + c 0 is a x + by + c 0 ANGLE BETWEEN TWO STRAIGHT LINES IN TERMS OF THEIR SLOPES : If m & m are the slopes of two itersectig straight lies (m m ) & θ is the acute agle betwee them, the Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 4 of 4

ta θ m m + m m Note : Let m, m, m are the slopes of three lies L 0 ; L 0 ; L 0 where m > m > m the the iterior agles of the ABC foud by these lies are give by, ta A m m ; ta B m m & ta C m m + m m + m m + m m PARALLEL LINES : (i) Whe two straight lies are parallel their slopes are equal Thus ay lie parallel to ax + by + c 0 is of the type ax + by + k 0 Where k is a parameter c c (ii) The distace betwee two parallel lies with equatios ax + by + c 0 & ax + by + c 0 is Note that the coefficiets of x & y i both the equatios must be same (iii) The area of the parallelogram p p siθ, where p & p are distaces betwee two pairs of opposite sides & θ is the agle betwee ay two adjacet sides Note that area of the parallelogram bouded by the lies y m x + c, ( c c) ( d d) y m x + c ad y m x + d, y m x + d is give by m m PERPENDICULAR LINES : (i) Whe two lies of slopes m & m are at right agles, the product of their slopes is, ie m m Thus ay lie perpedicular to ax + by + c 0 is of the form bx ay + k 0, where k is ay parameter (ii) Straight lies ax + by + c 0 & a x + b y + c 0 are at right agles if & oly if aa + bb 0 Equatios of straight lies through (x, y ) makig agle α with y mx + c are: (y y ) ta (θ α) (x x ) & (y y ) ta (θ + α) (x x ), where ta θ m 4 CONDITION OF CONCURRENCY : Three lies a x + b y + c 0, a x + b y + c 0 & a x + b y + c 0 are cocurret if a b c a b c 0 Alteratively : If three costats A, B & C ca be foud such that a b c A(a x + b y + c ) + B(a x + b y + c ) + C(a x + b y + c ) 0, the the three straight lies are cocurret 5 AREA OF A TRIANGLE : x y If (x i, y i ), i,, are the vertices of a triagle, the its area is equal to x y, provided the vertices are x y cosidered i the couter clockwise sese The above formula will give a ( ) ve area if the vertices (x i, y i ), i,, are placed i the clockwise sese 6 CONDITION OF COLLINEARITY OF THREE POINTS (AREA FORM): x y The poits (x i, y i ), i,, are colliear if x y x y 7 THE EQUATION OF A FAMILY OF STRAIGHT LINES PASSING THROUGH THE POINTS OF INTERSECTION OF TWO GIVEN LINES: The equatio of a family of lies passig through the poit of itersectio of a x + b y + c 0 & a x + b y + c 0 is give by (a x + b y + c ) + k(a x + b y + c ) 0, where k is a arbitrary real umber Note: If u ax + by + c, u a x + b y + d, u ax + by + c, u 4 a x + b y + d the, u 0; u 0; u 0; u 4 0 form a parallelogram u u u u 4 0 represets the diagoal BD Proof : Sice it is the first degree equatio i x & y it is a straight lie Secodly poit B satisfies the equatio because the co ordiates of B satisfy u 0 ad u 0 Similarly for the poit D Hece the result O the similar lies u u u u 4 0 represets the diagoal AC Note: The diagoal AC is also give by u + λu 4 0 ad u + µu 0, if the two equatios are idetical for some λ ad µ [For gettig the values of λ & µ compare the coefficiets of x, y & the costat terms] 8 BISECTORS OF THE ANGLES BETWEEN TWO LINES : (i) Equatios of the bisectors of agles betwee the lies ax + by + c 0 & a x + b y + c 0 (ab a b) are : a x + by + c a x + b y + c ± a + b a + b (ii) To discrimiate betwee the acute agle bisector & the obtuse agle bisector If θ be the agle betwee oe of the lies & oe of the bisectors, fid ta θ If ta θ <, the θ < 90 so that this bisector is the acute agle bisector If ta θ >, the we get the bisector to be the obtuse agle bisector (iii) To discrimiate betwee the bisector of the agle cotaiig the origi & that of the agle ot cotaiig the origi Rewrite the equatios, ax + by + c 0 & a x + b y + c 0 such that the costat terms c, c are positive The; a x + by + c a x + b y + c + gives the equatio of the bisector of the agle cotaiig the origi & a x + by + c a + b a x + b y + c gives the equatio of the bisector of the agle ot cotaiig the origi a + b (iv) To discrimiate betwee acute agle bisector & obtuse agle bisector proceed as follows Write ax + by + c 0 & a x + b y + c 0 such that costat terms are positive Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 5 of 4

If aa + bb < 0, the the agle betwee the lies that cotais the origi is acute ad the equatio of the bisector of ax+by+c a x + b y + c this acute agle is + a + b a + b therefore a x + by + c a x + b y + c is the equatio of other bisector a + b If, however, aa + bb > 0, the the agle betwee the lies that cotais the origi is obtuse & the equatio of the bisector of this obtuse agle is: a x + by + c a x + b y + c + ; therefore a x + by + c a x + b y + c a + b a + b is the equatio of other bisector (v) Aother way of idetifyig a acute ad obtuse agle bisector is as follows : Let L 0 & L 0 are the give lies & u 0 ad u 0 are the bisectors betwee L 0 & L 0 Take a poit P o ay oe of the lies L 0 or L 0 ad drop perpedicular o u 0 & u 0 as show If, p < q u is the acute agle bisector p > q u is the obtuse agle bisector p q the lies L & L are perpedicular Note : Equatio of straight lies passig through P(x, y ) & equally iclied with the lies a x + b y + c 0 & a x + b y + c 0 are those which are parallel to the bisectors betwee these two lies & passig through the poit P 9 A PAIR OF STRAIGHT LINES THROUGH ORIGIN : (i) A homogeeous equatio of degree two of the type ax² + hxy + by² 0 always represets a pair of straight lies passig through the origi & if : (a) h² > ab lies are real & distict (b) h² ab lies are coicidet (c) h² < ab lies are imagiary with real poit of itersectio ie (0, 0) (ii) If y m x & y m x be the two equatios represeted by ax² + hxy + by 0, the; m + m h & m b m a b (iii) If θ is the acute agle betwee the pair of straight lies represeted by, ax + hxy + by 0, the; ta θ h a b The coditio that these lies are: (a) At right agles to each other is 0 ie co efficiet of x + coefficiet of y 0 (b) Coicidet is h ab (c) Equally iclied to the axis of x is h 0 ie coeff of xy 0 Note: A homogeeous equatio of degree represets straight lies passig through origi 0 GENERAL EQUATION OF SECOND DEGREE REPRESENTING A PAIR OF STRAIGHT LINES: (i) ax + hxy + by + gx + fy + c 0 represets a pair of straight lies if: a h g abc + fgh af bg ch 0, ie if h b f 0 g f c (ii) The agle θ betwee the two lies represetig by a geeral equatio is the same as that betwee the two lies represeted by its homogeeous part oly The joit equatio of a pair of straight lies joiig origi to the poits of itersectio of the lie give by lx + my + 0 (i) & the d degree curve : ax² + hxy + by² + gx + fy + c 0 (ii) is ax + hxy + by lx+ my x+ my x+ my + gx fy c + l + l 0 (iii) lx+ my (iii) is obtaied by homogeizig (ii) with the help of (i), by writig (i) i the form: The equatio to the straight lies bisectig the agle betwee the straight lies, ax + hxy + by 0 is x y xy a b h The product of the perpediculars, dropped from (x, y ) to the pair of lies represeted by the equatio, ax² + a x + hxy + by hxy + by² 0 is ( a b) + 4 h 4 Ay secod degree curve through the four poit of itersectio of f(x y) 0 & xy 0 is give by f (x y) + λ xy 0 where f(xy) 0 is also a secod degree curve EXERCISE Q The sides AB, BC, CD, DA of a quadrilateral have the equatios x + y, x, x y 4, 5x + y + 0 respectively Fid the agle betwee the diagoals AC & BD Q Fid the co-ordiates of the orthocetre of the triagle, the equatios of whose sides are x + y, x + y 6, 4x y + 4 0, without fidig the co ordiates of its vertices Q Two vertices of a triagle are (4, ) & (, 5) If the orthocetre of the triagle is at (, ), fid the coordiates of the third vertex Q4 The poit A divides the joi of P ( 5, ) & Q (, 5) i the ratio K : Fid the two values of K for which the area of triagle ABC, where B is (, 5) & C is (7, ), is equal to uits i magitude Q5 Determie the ratio i which the poit P(, 5) divides the joi of A(, ) & B(7, 9) Fid the harmoic cojugate of P wrt A & B Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 6 of 4

Q6 A lie is such that its segmet betwee the straight lies 5x y 4 0 ad x + 4y 4 0 is bisected at the poit (, 5) Obtai the equatio Q7 A lie through the poit P(, ) meets the lies x y + 7 0 ad x + y 0 at the poits A ad B respectively If P divides AB exterally i the ratio : the fid the equatio of the lie AB Q8 The area of a triagle is 5 Two of its vertices are (, ) & (, ) The third vertex lies o y x + Fid the third vertex Q9 A variable lie, draw through the poit of itersectio of the straight lies x y + & x y +, meets the a b b a coordiate axes i A & B Show that the locus of the mid poit of AB is the curve xy() ab(x + y) Q0 Two cosecutive sides of a parallelogram are 4x + 5y 0 & 7x + y 0 If the equatio to oe diagoal is x + 7y 9, fid the equatio to the other diagoal Q The lie x + y 4 meets the y axis at A & the x axis at B The perpedicular bisector of AB meets the lie through (0, ) parallel to x axis at C Fid the area of the triagle ABC Q If the straight lie draw through the poit P (, ) & makig a agle π with the x axis, meets the lie x 6 4y + 8 0 at Q Fid the legth PQ Q Fid the coditio that the diagoals of the parallelogram formed by the lies ax + by + c 0; ax + by + c 0; a x + b y + c 0 & a x + b y + c 0 are at right agles Also fid the equatio to the diagoals of the parallelogram Q4 If lies be draw parallel to the axes of co-ordiates from the poits where x cosα + y siα p meets them so as to meet the perpedicular o this lie from the origi i the poits P ad Q the prove that PQ 4p cosα cosec α Q5 The poits (, ) & (5, ) are two opposite vertices of a rectagle The other two vertices lie o the lie y x + c Fid c & the remaiig vertices Q6 A straight lie L is perpedicular to the lie 5x y The area of the triagle formed by the lie L & the coordiate axes is 5 Fid the equatio of the lie Q7 Two equal sides of a isosceles triagle are give by the equatios 7x y + 0 ad x + y 0 & its third side passes through the poit (, 0) Determie the equatio of the third side Q8 The vertices of a triagle OBC are O (0, 0), B (, ), C(, ) Fid the equatio of the lie parallel to BC & itersectig the sides OB & OC, whose perpedicular distace from the poit (0, 0) is half Q9 Fid the directio i which a straight lie may be draw through the poit (, ) so that its poit of itersectio with the lie 4y 4x + 4 + + 0 0 is at a distace of uits from (, ) Q0 Cosider the family of lies, 5x + y + K (x y 4) 0 ad x y + + K (x y )0 Fid the equatio of the lie belogig to both the families without determiig their vertices Q Give vertices A (, ), B (4, ) & C (5, 5) of a triagle, fid the equatio of the perpedicular dropped from C to the iterior bisector of the agle A Q If through the agular poits of a triagle straight lies be draw parallel to the opposite sides, ad if the itersectios of these lies be joied to the opposite agular poits of the traigle the usig co-ordiate geometry, show that the lies so obtaied are cocurret Q Determie all values of α for which the poit (α, α²) lies iside the triagle formed by the lies x + y 0 ; x + y 0 ; 5x 6y 0 Q4 If the equatio, ax + hxy + by + gx + fy + c 0 represet a pair of straight lies, prove that the equatio to the third pair of straight lies passig through the poits where these meet the axes is, ax hxy + by + gx + fy + c + 4fg xy 0 c Q5 A straight lie is draw from the poit (, 0) to the curve x + y + 6x 0y + 0, such that the itercept made o it by the curve subteds a right agle at the origi Fid the equatios of the lie Q6 Determie the rage of values of θ [0, π] for which the poit (cos θ, si θ) lies iside the triagle formed by the lies x + y ; x y & 6x + y 0 0 Q7 Fid the co-ordiates of the icetre of the triagle formed by the lie x + y + 0; x y + 0 & 7x y + 0 Also fid the cetre of the circle escribed to 7x y + 0 B D A B Q8 I a triagle ABC, D is a poit o BC such that The equatio of the lie AD is DC AC x + y + 4 0 & the equatio of the lie AB is x + y + 0 Fid the equatio of the lie AC Q9 Show that all the chords of the curve x y x + 4y 0 which subted a right agle at the origi are cocurret Does this result also hold for the curve, x² + y² x + 4y 0? If yes, what is the poit of cocurrecy & if ot, give reasos Q0 Without fidig the vertices or agles of the triagle, show that the three straight lies au + bv 0; au bv ab ad u + b 0 from a isosceles triagle where u x + y b & v x y a & a, b 0 EXERCISE Q The equatios of perpediculars of the sides AB & AC of triagle ABC are x y 4 0 ad 5 x y 5 0 respectively If the vertex A is (, ) ad poit of itersectio of perpediculars bisectors is,, fid the equatio of medias to the sides AB & AC respectively Q A lie 4x + y through the poit A(, 7) meets the lie BC whose equatio is x 4y + 0 at a poit B Fid the equatio of the lie AC, so that AB AC α Q If x cos α + y si α p, where p si be a straight lie, prove that the perpediculars o this straight lie from cosα the poits (m², m), (mm, m + m ), (m ², m ) form a GP Q4 A(, 0) ad B(6, 0) are two fixed poits ad P(x, y ) is a variable poit AP ad BP meet the y-axis at C & D respectively ad AD meets OP at Q where 'O' is the origi Prove that CQ passes through a fixed poit ad fid its co ordiates Q5 Fid the equatio of the straight lies passig through (, 7) & havig a itercept of legth betwee the straight lies 4x + y, 4x + y Q6 Let ABC be a triagle with AB AC If D is the mid poit of BC, E the foot of the perpedicular from D to AC ad F the midpoit of DE, prove aalytically that AF is perpedicular to BE Q7 Two sides of a rhombous ABCD are parallel to the lies y x + & y 7x + If the diagoals of the rhombous itersect at the poit (, ) & the vertex A is o the y-axis, fid the possible coordiates of A Teko Classes, Maths : Suhag R Kariya (S R K Sir), Bhopal, Phoe : 0 90 90 777 9, 0 9890 5888 page 7 of 4