1 / 65

Mathematics

Mathematics. Session. Cartesian Coordinate Geometry And Straight Lines. Session Objectives. Session Objectives. Equations of bisectors of angles between two lines Acute/obtuse angle bisectors Position of origin w.r.t bisecors Equation of family of lines through intersection of two lines

nathaniel-carney
Download Presentation

Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mathematics

  2. Session Cartesian Coordinate Geometry And Straight Lines

  3. Session Objectives

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

  5. Y a2x+b2y+c2=0 N A P(h,k) a1x+b1y+c1=0 M X’ O X Y’ Equation of Bisector Consider two lines a1x+b1y+c1 = 0 and a2x+b2y+c2 = 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. Y a2x+b2y+c2=0 N A P(h,k) a1x+b1y+c1=0 M X’ O X Y’ Equation of Bisector 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 lines in general form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 such that c1 and c2 are positive. Step II : Determine sign of expression a1a2+b1b2 Step III : Write the equations of the bisectors ;

  8. Acute/obtuse Angle Bisectors Step IV : Case (i) a1a2+b1b2 > 0 Obtuse angle bisector Acute angle bisector

  9. Acute/obtuse Angle Bisectors Step IV : Case (i) a1a2+b1b2 < 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. Solution : Rewrite the equations to make the constant terms positive, 12x-5y+7 = 0 and 4x-3y+1 = 0 Calculate a1a2+b1b2 12*4+(-5)*(-3) = 63

  11. Solution Cont. a1a2+b1b2 > 0, therefore the obtuse angle bisector is Simplifying, Which is the required equation of the obtuse angle bisector.

  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 lines in general form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 such that c1 and c2 are positive. Step II : Determine sign of expression a1a2+b1b2

  13. Origin w.r.t. Angle Bisectors Step III : Case (i) a1a2+b1b2 > 0 Origin lies in obtuse angle between the lines

  14. Origin w.r.t. Angle Bisectors Step III : Case (i) a1a2+b1b2 < 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. Solution : Rewrite the equations to make the constant terms positive, -4x-3y+6 = 0 and 5x+12y+9 = 0 Calculate a1a2+b1b2 (-4)*5+(-3)*(12) = -56 a1a2+b1b2 < 0, therefore origin lies in the acute angle.

  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 a1x+b1y+c1 = 0 and a2x+b2y+c2 = 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. Solution : 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

  19. Pair of Lines - Locus Definition A pair of straight lines is the locus of a point whose coordinates satisfy a second degree equation ax2+2hxy+by2+2gx+2fy+c = 0 such that it can be factorized into two linear equations.

  20. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0 in general represents all the conics in the x-y plane.

  21. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0  = 0, h2-ab  0 A pair of lines

  22. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0   0, h2-ab > 0 A hyperbola

  23. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0   0, h2-ab = 0 A parabola

  24. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0   0, h2-ab < 0 An ellipse

  25. Pair of Lines ax2+2hxy+by2+2gx+2fy+c = 0   0, h = 0, a = b A circle

  26. Individual Lines To find the equation of individual lines in the pair of lines ax2+2hxy+by2+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 2x2-xy-y2+9x-3y+10 = 0.

  28. Solution Rewriting the given equation, 2x2-(y-9)x-(y2+3y-10) = 0 Which are the required equations.

  29. Point of Intersection Point of intersection of the pair of lines represented by ax2+2hxy+by2+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 nth 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 ax2+2hxy+by2 = 0 ax2+2hxy+by2 = 0 can be rewritten as b(y-m1x)(y-m2x) = 0, where m1 and m2 are the slopes of the two lines.  bm1m2x2-b(m1+m2)xy+by2 = 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 by 4x2+24xy+11y2 = 0. Solution :

  33. Pair of Lines Through Origin If a pair of straight lines is represented by ax2+2hxy+by2+2gx+2fy+c = 0, then ax2+2hxy+by2 = 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 ax2+2hxy+by2+2gx+2fy+c = 0, Method II : Step I : Find slopes of the individual lines m1 and m2 using ax2+2hxy+by2 = 0 Step II : compare coefficients in the identity ax2+2hxy+by2+2gx+2fy+c  b(y-m1x-c1)(y-m2x-c2) to find c1 and c2.

  35. Illustrative Example Find the separate equations of the lines represented by 2x2+5xy+3y2+6x+7y+4 = 0. Solution :

  36. Solution Cont. Consider the identity

  37. Y X’ O X Y’ Pair of Lines With Given Condition To find the joint equation of a pair of lines joining the origin to the points of intersection of the curve ax2+2hxy+by2+2gx+2fy+c = 0 and the line lx+my+n = 0.

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

  39. Y X’ O X Y’ Pair of Lines With Given Condition 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 x2+2xy+3y2+4x+8y-11 = 0. Solution : Equation of straight line can be rewritten as Using this to make the equation of the curve homogeneous,

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

  42. Angle Between a Pair of Lines If ax2+2hxy+by2+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 ax2+2hxy+by2+2gx+2fy+c = 0 represents a pair of lines, Acute angle between the lines is given by : Lines are parallel or coincident if h2 = ab

  44. Angle Between a Pair of Lines If ax2+2hxy+by2+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. Acute angle between the pair of lines is given by Illustrative Example Find the angle between the pair of lines represented by 2x2+5xy+3y2+6x+7y+4 = 0. Solution :

  46. Angle Bisectors of a Pair of Lines If ax2+2hxy+by2 = 0 represents a pair of lines, Equation of angle bisectors is given by :

  47. Perpendiculars to a Pair of Lines If ax2+2hxy+by2 = 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 135x2-136xy+33y2 = 0. Solution : Equation of bisectors are given by : Which is the required equation.

  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. Solution : The equations can be rewritten as Angle bisectors are given by

More Related