Mathematics

1. Mathematics

2. Session Cartesian Coordinate Geometry And Straight Lines

3. Session Objectives • Cartesian Coordinate system and Quadrants • Distance formula • Area of a triangle • Collinearity of three points • Section formula • Special points in a triangle • Locus and equation to a locus • Translation of axes - shift of origin • Translation of axes - rotation of axes

4. René Descartes

5. Y 3 2 +ve direction 1 X’ O X -4 -3 -2 -1 2 3 4 1 -1 Origin +ve direction -ve direction -2 -ve direction -3 Y’ Coordinates Y-axis : Y’OY X-axis : X’OX

6. Y 3 2 1 X’ O X -4 -3 -2 -1 2 3 4 1 -1 -2 (?,?) -3 Y’ Coordinates (2,1) Abcissa Ordinate (-3,-2)

7. Y 3 2 1 X’ O X -4 -3 -2 -1 2 3 4 1 -1 -2 -3 Y’ Coordinates (2,1) Abcissa Ordinate (-3,-2) (4,?)

8. Y 3 2 1 X’ O X -4 -3 -2 -1 2 3 4 1 -1 -2 -3 Y’ Coordinates (2,1) Abcissa Ordinate (-3,-2) (4,-2.5)

9. Y X’ O X Y’ Quadrants (-,+) (+,+) II I IV III (+,-) (-,-)

10. Y (-,+) (+,+) II I X’ O X IV III (+,-) (-,-) Y’ Quadrants Ist? IInd? Q : (1,0) lies in which Quadrant? A : None. Points which lie on the axes do not lie in any quadrant.

11. Y Q(x2, y2) y2-y1 y2 N y1 P(x1, y1) x2 x1 (x2-x1) X’ O X Y’ Distance Formula PQN is a right angled .  PQ2 = PN2 + QN2  PQ2 = (x2-x1)2+(y2-y1)2

12. Distance From Origin Distance of P(x, y) from the origin is

13. Applications of Distance Formula Parallelogram

14. Applications of Distance Formula Rhombus

15. Applications of Distance Formula Rectangle

16. Applications of Distance Formula Square

17. A(x1, y1) Y C(x3, y3) B(x2, y2) M L N X’ O X Y’ Area of a Triangle Area of  ABC = Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC

18. A(x1, y1) Y C(x3, y3) B(x2, y2) M L N X’ O X Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC Y’ Area of a Triangle Sign of Area : Points anticlockwise  +ve Points clockwise  -ve

19. Area of Polygons Area of polygon with points Ai (xi, yi) where i = 1 to n Can be used to calculate area of Quadrilateral, Pentagon, Hexagon etc.

20. Collinearity of Three Points Method I : Use Distance Formula a b c Show that a+b = c

21. Collinearity of Three Points Method II : Use Area of Triangle A  (x1, y1) B  (x2, y2) C  (x3, y3) Show that

22. Y B(x2, y2) n P(x, y) : K m H A(x1, y1) L N M X’ O X Y’ Section Formula – Internal Division Clearly AHP ~ PKB

23. Midpoint Midpoint of A(x1, y1) and B(x2,y2) m:n  1:1

24. Y P(x, y) B(x2, y2) K H A(x1, y1) L N M X’ O X Y’ Section Formula – External Division P divides AB externally in ratio m:n Clearly PAH ~ PBK

25. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Intersection of medians of a triangle is called the centroid. Centroid is always denoted by G.

26. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

27. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

28. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

29. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Medians are concurrent at the centroid, centroid divides medians in ratio 2:1 We see that L  M  N  G

30. A(x1, y1) F E G B(x2, y2) D C(x3, y3) Centroid Centroid We see that L  M  N  G

31. A(x1, y1) F E I B(x2, y2) D C(x3, y3) Incentre Intersection of angle bisectors of a triangle is called the incentre Incentre is the centre of the incircle Let BC = a, AC = b, AB = c AD, BE and CF are the angle bisectors of A, B and C respectively.

32. A(x1, y1) F E I B(x2, y2) D C(x3, y3) Incentre Similarly I can be derived using E and F also

33. A(x1, y1) F E I B(x2, y2) D C(x3, y3) Incentre Angle bisectors are concurrent at the incentre

34. E A(x1, y1) F E B(x2, y2) D C(x3, y3) EA = Excentre opposite A Excentre Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle

35. E A(x1, y1) F E B(x2, y2) D C(x3, y3) EB = Excentre opposite B Excentre Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle

36. E A(x1, y1) F E B(x2, y2) D C(x3, y3) EC = Excentre opposite C Excentre Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle

37. A C O B Cirumcentre Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre. OA = OB = OC = circumradius The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.

38. A H B C Orthocentre Intersection of altitudes of a triangle is called the orthocentre. Orthocentre is always denoted by H We will learn to find coordinates of Orthocentre after we learn straight lines and their equations

39. H 1 : 2 G O Cirumcentre, Centroid and Orthocentre The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear. G divides OH in the ratio 1:2

40. Locus – a Definition The curve described by a point which moves under a given condition or conditions is called its locus e.g. locus of a point having a constant distance from a fixed point : Circle!!

41. Locus – a Definition The curve described by a point which moves under a given condition or conditions is called its locus e.g. locus of a point equidistant from two fixed points : Perpendicular bisector!!

42. Equation to a Locus The equation to the locus of a point is that relation which is satisfied by the coordinates of every point on the locus of that point Important : A Locus is NOT an equation. But it is associated with an equation

43. Equation to a Locus Algorithm to find the equation to a locus : Step I : Assume the coordinates of the point whose locus is to be found to be (h,k) Step II : Write the given conditions in mathematical form using h, k Step III : Eliminate the variables, if any Step IV : Replace h by x and k by y in Step III. The equation thus obtained is the required equation to locus

44. Illustrative Example Find the equation to the locus of the point equidistant from A(1, 3) and B(-2, 1) Solution : Let the point be P(h,k) PA = PB (given)  PA2 = PB2  (h-1)2+(k-3)2 = (h+2)2+(k-1)2  6h+4k = 5  equation of locus of (h,k) is 6x+4y = 5

45. B(0,b) P(h,k) O A(a,0) Illustrative Example A rod of length l slides with its ends on perpendicular lines. Find the locus of its midpoint. Solution : Let the point be P(h,k) Let the  lines be the axes Let the rod meet the axes at A(a,0) and B(0,b)  h = a/2, k = b/2 Also, a2+b2 = l2  4h2+4k2 = l2  equation of locus of (h,k) is 4x2+4y2 = l2

46. Y P(x,y) Y X y O’(h,k) x X’ O X Y’ Shift of Origin Consider a point P(x, y) Let the origin be shifted to O’ with coordinates (h, k) relative to old axes Let new P  (X, Y)  x = X + h, y = Y + k  X = x - h, Y = y - k O  (-h, -k) with reference to new axes

47. Illustrative Problem Show that the distance between two points is invariant under translation of the axes Solution : Let the points have vertices A(x1, y1), B(x2, y2) Let the origin be shifted to (h, k) new coordinates : A(x1-h, y1-k), B(x2-h, y2-k) = Old dist.

48. Y Y Y P(x,y) Y Y Y X X X y X X X’ O X O O O O X’ x X’ Y’ Y’ Y’ Y’ Y’ X’ X’ Rotation of Axes R Consider a point P(x, y)  Let the axes be rotated through an angle .  Let new P  (X, Y) make an angle  with the new x-axis

49. Rotation of Axes

50. Class Exercise