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# Mathematics by SHAMS KHAN

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1. Mathematics by SHAMS KHAN

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 Rectangle

15. 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

16. 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

17. 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.

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

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

20. 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

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

22. 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

23. 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.

24. 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

25. 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

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 Medians are concurrent at the centroid, centroid divides medians in ratio 2:1 We see that L  M  N  G

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

29. 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.

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

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

32. 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

33. 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

34. 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

35. 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.

36. 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

37. 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

38. 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!!

39. 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!!

40. 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

41. 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

42. 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

43. 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

44. 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

45. 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.

46. 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

47. Rotation of Axes

48. Class Exercise