1. Mathematics

2. Cartesian Coordinate Geometry And Straight Lines Session

3. Session Objectives

4. 1.Equations of bisectors of angles between two lines 2.Acute/obtuse angle bisectors 3.Position of origin w.r.t bisecors 4.Equation of family of lines through intersection of two lines 5.Pair of lines - locus definition 6.Pair of lines represented by second degree equation 7.Angle between two lines, represented as a second degree equation Session Objectives

5. Equation of Bisector XX’ Y’ O Y a 1 x+b 1 y+c 1 =0 a 2 x+b 2 y+c 2 =0 AM N P(h,k) Consider two lines a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 We are required to find the equations of the bisectors of the angle between them. The required equations are the equations to the locus of a point P(h, k) equidistant from the given lines.

6. Equation of Bisector XX’ Y’ O Y a 1 x+b 1 y+c 1 =0 a 2 x+b 2 y+c 2 =0 AM N P(h,k) PM = PN  The required equations are

7. Acute/obtuse Angle Bisectors Algorithm to determine equations of bisectors of acute angle and obtuse angle between a pair of lines. Step I : Rewrite the equations of the linesin general form a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 such that c 1 and c 2 are positive. Step II : Determine sign of expression a 1 a 2 +b 1 b 2 Step III : Write the equations of the bisectors ;

8. Acute/obtuse Angle Bisectors Step IV : Case (i) a 1 a 2 +b 1 b 2 > 0 Obtuse angle bisector Acute angle bisector

9. Acute/obtuse Angle Bisectors Step IV : Case (i) a 1 a 2 +b 1 b 2 < 0 Acute angle bisector Obtuse angle bisector

10. Illustrative Example Find the equation of the obtuse angle bisector of lines 12x-5y+7 = 0 and 3y-4x-1 = 0. Rewrite the equations to make the constant terms positive, 12x-5y+7 = 0 and 4x-3y+1 = 0 Calculate a 1 a 2 +b 1 b 2 12*4+(-5)*(-3) = 63 Solution :

11. Solution Cont. Simplifying, Which is the required equation of the obtuse angle bisector. a 1 a 2 +b 1 b 2 > 0, therefore the obtuse angle bisector is

12. Origin w.r.t. Angle Bisectors Algorithm to determine whether origin lies in the obtuse angle or the acute angle between a pair of lines Step I : Rewrite the equations of the linesin general form a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 such that c 1 and c 2 are positive. Step II : Determine sign of expression a 1 a 2 +b 1 b 2

13. Origin w.r.t. Angle Bisectors Step III : Case (i) a 1 a 2 +b 1 b 2 > 0 Origin lies in obtuse angle between the lines

14. Origin w.r.t. Angle Bisectors Step III : Case (i) a 1 a 2 +b 1 b 2 < 0 Origin lies in acute angle between the lines

15. Illustrative Example For the straight lines 4x+3y-6 = 0 and 5x+12y+9 = 0 find the equation of the bisector of the angle which contains the origin. Rewrite the equations to make the constant terms positive, -4x-3y+6 = 0 and 5x+12y+9 = 0 Calculate a 1 a 2 +b 1 b 2 (-4)*5+(-3)*(12) = -56 a 1 a 2 +b 1 b 2 < 0, therefore origin lies in the acute angle. Solution :

16. Illustrative Example Acute angle bisector is given by :  The origin lies in the acute angle and the equation of the acute angle bisector is 7x+9y-3 = 0.

17. Family of Lines Through Intersection of a Pair of Lines Equation of the family of lines passing through the intersection of the lines a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 is given by can be calculated using some given condition

18. Illustrative Example Find the equation of the straight line which passes through the point (2, -3) and the point of intersection of x+y+4 = 0 and 3x-y-8 = 0. Required equation can be written as x+y+4+(3x-y-8) = 0 where is a parameter. This passes through (2, -3).  2-3+4+ (3*2+3-8) = 0.  = -3  the required equation is x+y+4-3(3x-y-8) = 0 or –8x+4y+28 = 0 or 2x-y-7 = 0 Solution :

19. Pair of Lines - Locus Definition A pair of straight lines is the locus of a point whose coordinates satisfy a second degree equation ax 2 +2hxy+by 2 +2gx+2fy+c = 0 such that it can be factorized into two linear equations.

20. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 in general represents all the conics in the x-y plane.

21. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 A pair of lines  = 0, h 2 -ab  0

22. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 A hyperbola   0, h 2 -ab > 0

23. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 A parabola   0, h 2 -ab = 0

24. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 An ellipse   0, h 2 -ab < 0

25. Pair of Lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0 A circle   0, h = 0, a = b

26. Individual Lines To find the equation of individual lines in the pair of lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0, Method I : Step I : Rewrite the equation as a quadratic in x (or y). Step II : Solve for x (or y). Method II : Will be discussed later.

27. Illustrative Example Find the separate equations of the lines represented by 2x 2 -xy-y 2 +9x-3y+10 = 0.

28. Solution Rewriting the given equation, 2x 2 -(y-9)x-(y 2 +3y-10) = 0 Which are the required equations.

29. Point of Intersection Point of intersection of the pair of lines represented by ax 2 +2hxy+by 2 +2gx+2fy+c = 0 is No need to memories this formula. To find the required point, find the equations of the individual lines and solve simultaneously

30. Homogeneous Equation An equation, with R.H.S. 0, in which the sum of the powers of x and y in every term is the same, say n, is called a homogeneous equation of n th degree in x and y.

31. Pair of Lines Through Origin A pair of straight lines passing through the origin is represented by a homogeneous equation of second degree ax 2 +2hxy+by 2 = 0 ax 2 +2hxy+by 2 = 0 can be rewritten as b(y-m 1 x)(y-m 2 x) = 0, where m 1 and m 2 are the slopes of the two lines.  bm 1 m 2 x 2 -b(m 1 +m 2 )xy+by 2 = 0 The above relations can be used to find the equations and the slopes of the individual lines.

32. Illustrative Example Find the separate equations of the lines represented by4x 2 +24xy+11y 2 = 0. Solution :

33. Pair of Lines Through Origin If a pair of straight lines is represented by ax 2 +2hxy+by 2 +2gx+2fy+c = 0, then ax 2 +2hxy+by 2 = 0 represents a pair of lines parallel to them and passing through the origin.

34. Individual Lines To find the equation of individual lines in the pair of lines ax 2 +2hxy+by 2 +2gx+2fy+c = 0, Method II : Step I : Find slopes of the individual lines m 1 and m 2 using ax 2 +2hxy+by 2 = 0 Step II : compare coefficients in the identity ax 2 +2hxy+by 2 +2gx+2fy+c  b(y-m 1 x-c 1 )(y-m 2 x-c 2 ) to find c 1 and c 2.

35. Illustrative Example Find the separate equations of the lines represented by 2x 2 +5xy+3y 2 +6x+7y+4 = 0. Solution :

36. Solution Cont. Consider the identity

37. Pair of Lines With Given Condition XX’ Y’ O Y To find the joint equation of a pair of lines joining the origin to the points of intersection of the curve ax 2 +2hxy+by 2 +2gx+2fy+c = 0 and the line lx+my+n = 0.

38. Pair of Lines With Given Condition XX’ Y’ O Y Step I : Rewrite lx+my+n = 0 such that R.H.S. = 1 Step II : Make the equation of the curve homogeneous

39. Pair of Lines With Given Condition XX’ Y’ O Y The required equation is the equation arrived at in Step II.

40. Illustrative Example Find the equation of the lines joining the origin to the points of intersection of the straight line y = 3x+2 and the curver x 2 +2xy+3y 2 +4x+8y-11 = 0. Equation of straight line can be rewritten as Using this to make the equation of the curve homogeneous, Solution :

41. Solution On simplifying, Which is the required equation.

42. Angle Between a Pair of Lines If ax 2 +2hxy+by 2 +2gx+2fy+c = 0 represents a pair of lines, Acute angle between the lines is given by : Independent of g, f, c

43. Angle Between a Pair of Lines If ax 2 +2hxy+by 2 +2gx+2fy+c = 0 represents a pair of lines, Acute angle between the lines is given by : Lines are parallel or coincident if h 2 = ab

44. Angle Between a Pair of Lines If ax 2 +2hxy+by 2 +2gx+2fy+c = 0 represents a pair of lines, Acute angle between the lines is given by : Lines are perpendicular if a+b = 0

45. Illustrative Example Find the angle between the pair of lines represented by 2x 2 +5xy+3y 2 +6x+7y+4 = 0. Acute angle between the pair of lines is given by Solution :

46. Angle Bisectors of a Pair of Lines If ax 2 +2hxy+by 2 = 0 represents a pair of lines, Equation of angle bisectors is given by :

47. Perpendiculars to a Pair of Lines If ax 2 +2hxy+by 2 = 0 represents a pair of lines, Equation of perpendiculars to the pair of lines is given by :

48. Illustrative Example Find the equation of the bisectors of the lines represented by 135x 2 -136xy+33y 2 = 0. Equation of bisectors are given by : Which is the required equation. Solution :

49. Class Exercise

50. Class Exercise - 1 The bisectors of the angle between the lines y = 3x+3 and 3y = x+33 meet the X-axis at P and Q. Find length PQ. The equations can be rewritten as Angle bisectors are given by Solution :

51. Solution Cont. These lines meet the X-axis at P(3,0) and Q(-3,0). Clearly length PQ = 6.

52. Class Exercise - 2 Prove that the bisectors of the angles formed by any two intersecting lines are perpendicular to each other. A.Consider two intersecting lines Consider the angle bisectors Q.E.D. +++ =   + = /2     Solution :

53. Class Exercise - 3 Show that the reflection of the lines px+qy+r = 0 in the line x+y+1 = 0 is the line qx+py+(p+q-r) = 0, where p  -q Angle bisectors of the lines px+qy+r = 0 and qx+py+(p+q-r) = 0 are Second equation can be written as x+y+1 = 0 (as p+q  0) Q.E.D. Solution :

54. Class Exercise - 4 A rhombus has two of its sides parallel to the lines y = 2x+3 and y = 7x+2. If the diagonals cut at (1,2) and one vertex is on the Y-axis, find the possible values of the ordinate of that vertex. Concept : Diagonals of a rhombus bisect the angle. Let the sides intersecting at the Y-axis be 2x-y+ 1 = 0 and 7x-y+ 2 = 0. These lines will meet the Y-axis at (0, 1 ) and (0, 2 ).  1 = 2 = (say) Thus the sides are 2x-y+ = 0 and 7x-y+ = 0. Solution :

55. Angle bisectors will be These will pass through (1,2) Solution Cont.

56. Class Exercise - 5 Find the bisector of the angle between the lines 2x+y-6 = 0 and 2x-4y+7 = 0 which contains the point (1,2). Consider a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 (c 1, c 2  0), If a 1 h+b 1 k+c 1 and a 2 h+b 2 k+c 2 have the same sign, point (h,k) will lie in the angle bisector Solution :

57. Angle bisector containing (1,2) is Solution Cont. Given lines can be rewritten as -2x-y+6 = 0 and 2x-4y+7 = 0. Now -2(1)-(2)+6 > 0 and 2(1)-4(2)+7 > 0

58. Class Exercise - 6 Find the equation of the straight line drawn through the point of intersection of the lines x+y = 4 and 2x-3y = 1 and perpendicular to the line cutting off intercepts 5 and 6 on the positive axes. Family of lines through the intersection of the given lines is (x+y-4)+(2x-3y-1) = 0 Line cutting intercepts of 5 and 6 on the positive axes is Solution :

59. Thus required equation is 3(x+y-4)+11(2x-3y-1) = 0 Or, 25x-30y-23 = 0 Solution Cont. Slope of line perpendicular to this line will be -5/6.

60. Class Exercise - 7 Find the separate equations of the pair of lines represented by 12x 2 +5xy-28y 2 +19x+61y-21 = 0. Rewrite the equation as a quadratic in x, we get12x 2 +(5y+19)x-(28y 2 -61y+21) = 0 Solution :

61. Class Exercise - 8 Show that the lines joining the origin to the points common to x 2 +hxy-y 2 +gx+fy = 0 and fx-gy = are at right angles whatever be the value of. Given line can be rewritten as Using this to make the equation of the curve homogeneous, we get Solution :

62. Sum of coefficients of x 2 and y 2 = (+fg)-(+fg) = 0 Thus the required lines are perpendicular to each other, whatever be the value of. Q.E.D. Solution Cont.

63. Class Exercise - 9 Find the angle between the pair of lines : (x 2 +y 2 )sin 2  = (xcos-ysin) 2. Given equation can be rewritten as (cos 2 -sin 2 )x 2 -2sincosxy+sin 2 -sin 2 )y 2 = 0 Solution :

64. Class Exercise - 10 Prove that the lines a 2 x 2 +2h(a+b)xy+b 2 y 2 = 0 are equally inclined to the lines ax 2 +2hxy+by 2 = 0.

65. Thank you