1. Mathematics

2. Cartesian Coordinate Geometry
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
Cartesian coordinate geometry is so named in honour of Rene Descartes. Rene Descartes, considered by many to be the father of modern philosophy was also an outstanding Mathematician. It is said that he invented cartesian coordinate geometry on observing a fly walk across the ceiling. Descartes used his method to apply principles of algebra to geometrical problems.

5. Coordinates Y Y’ 1 2 3 X X’ O Origin -1 -2 -3 -4 1 2 3 4 -1 -2 -3
+ve direction
Y-axis : Y’OY
X-axis : X’OX X X’ O
Origin -1 -2 -3 -4
-ve direction 1 2 3 4
+ve direction -1 -2 -3
-ve direction
Description of axes. The axes are arbitrarily chosen mutually perpendicular lines passing through an arbitrarily chosen point identified as the origin. Left to right and bottom to top are positive directions.

6. Coordinates X X’ Y Y’ O 1 2 3 4 -1 -2 -3 -4 (2,1) Abcissa Ordinate
Any point on the plane can be uniquely identified by its x-coordinate and y-coordinate. The x-coordinate is also called the abcissa, the y coordinate the ordinate of the point. The point is identified as an ordered pair (abcissa, ordinate) or (x,y).
Ordinate
(-3,-2)
(?,?)

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

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

9. II I III IV (-,+) (+,+) (-,-) (+,-) Quadrants X X’ O Y Y’
The cartesian plane is divided into four quadrants. Quadrants are numbered in anticlockwise direction. All abcissae in a given quadrant will have the same sign and all ordinates in a given quadrant will have the same sign.
(-,-)
(+,-)

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

11. Distance Formula PQN is a right angled .  PQ2 = PN2 + QN2X X’ Y’ O Y
Q(x2, y2) N
y2-y1 y2
P(x1, y1)
PQN is a right angled .
 PQ2 = PN2 + QN2 y1 x1
(x2-x1)
 PQ2 = (x2-x1)2+(y2-y1)2
Explain the derivation step by step according to animation. x2

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

13. Applications of Distance Formula
Parallelogram
To show that a figure is a parallelogram, prove that opposite sides are equal.

14. Applications of Distance Formula
Rhombus
To show that a figure is a rhombus, prove all sides are equal.

15. Applications of Distance Formula
Rectangle
To show that a figure is a rectangle, prove that opposite sides are equal and the diagonals are equal. Corollary : Parellelogram but not rectangle  diagonals are unequal

16. Applications of Distance Formula
Square
To show that a figure is a square, prove all sides are equal and the diagonals are equal. Corollary : Rhombus but not square  diagonals are unequal

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

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

19. Can be used to calculate area of Quadrilateral, Pentagon, Hexagon etc.
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
To show that three points are collinear, show that a+b = c 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
To show that three points are collinear, show that area of triangle is zero

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

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

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

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

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

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

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

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

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

31. Incentre is the centre of the incircle
Intersection of angle bisectors of a triangle is called the incentre
A(x1, y1)
B(x2, y2)
C(x3, y3) D E F I
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. Similarly I can be derived using E and F also
Incentre
A(x1, y1)
B(x2, y2)
C(x3, y3) D E F I
Similarly I can be derived using E and F also

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

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

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

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

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

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

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

40. Circle!! 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. A Locus is NOT an equation. But it is associated with an equation
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
No variable is involved

45. 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
B(0,b)
A(a,0) O
P(h,k)
No variable is involved

46. Shift of Origin Consider a point P(x, y)
O’(h,k)
P(x,y) x y X Y
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. Rotation of Axes 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
Explain when to use which

50. Class Exercise

51. Class Exercise - 1
If the segments joining the points A(a,b) and B(c,d) subtend an angle  at the origin, prove that

52. Solution Let O be the origin.
 OA2 = a2+b2, OB2 = c2+d2, AB2 = (c-a)2+(d-b)2
Using Cosine formula in OAB, we have
AB2 = OA2+OB2-2OA.OBcos
On simplifying,

53. Class Exercise - 2
Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given such that
Find x.
Solution :
Given that ABC = 2DBC

54. Class Exercise - 3
If a  b  c, prove that (a,a2), (b,b2) and (c,c2) can never be collinear.
Solution :
Let, if possible, the three points be collinear.
R2  R2-R1, R3  R3- R2
This method of proof is called reductio ad absurdum.

55. Solution Cont.
R2  R2-R3
This is possible only if a = b or b = c or c = a.
But a  b  c. Thus the points can never be collinear.
This method of proof is called reductio ad absurdum.
Q.E.D.

56. Class Exercise - 4
Three vertices of a parallelogram taken in order are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b). Find the fourth vertex.
Solution :
Let the fourth vertex be (x,y).
Diagonals bisect each other.
 the required vertex is (-b,b)

57. Class Exercise - 5
If G be the centroid of ABC and P be any point in the plane, prove that PA2+PB2+PC2=GA2+GB2+GC2+3GP2.
Solution :
Choose a coordinate system such that G is the origin and P lies along the X-axis.
Let A  (x1,y1), B  (x2,y2), C  (x3,y3), P  (p,0)
 LHS = (x1-p)2+y12+(x2-p)2+y22+(x3-p)2+y32
= (x12+y12)+(x22+y22)+(x32+y32)+3p2-2p(x1+x2+x3)
=GA2+GB2+GC2+3GP2
=RHS
Choosing the coordinate system in this manner does not affect the generality of the solution. x1+x2+x3=0 since G is the origin.
Q.E.D.

58. Class Exercise - 6
The locus of the midpoint of the portion intercepted between the axes by the line xcos+ysin = p, where p is a constant, is
note alpha is a variable and hence to be eliminated

59. Solution
Let the line intercept at the axes at A and B. Let R(h,k) be the midpoint of AB.
 Ans : (b)

60. Class Exercise - 7
A point moves so that the ratio of its distance from (-a,0) to (a,0) is 2:3. Find the equation of its locus.
Solution :
Let the point be P(h,k). Given that

61. Class Exercise - 8
Find the locus of the point such that the line segments having end points (2,0) and (-2,0) subtend a right angle at that point.
Solution :
Let A  (2,0), B  (-2,0)
Let the point be P(h,k). Given that

62. Class Exercise - 9
Find the coordinates of a point where the origin should be shifted so that the equation x2+y2-6x+8y-9 = 0 will not contain terms in x and y. Find the transformed equation.
Solution :
Let the origin be shifted to (h,k). The given equation becomes (X+h)2+(Y+k)2-6(X+h)+8(Y+k)-9 = 0
Or, X2+Y2+(2h-6)X+(2k+8)Y+(h2+k2-6h+8k-9) = 0
 2h-6 = 0; 2k+8 = 0  h = 3, k = -4.
Thus the origin is shifted to (3,-4).
Transformed equation is X2+Y2+( ) = 0
Or, X2+Y2 = 34

63. Class Exercise - 10
Through what angle should the axes be rotated so that the equation 11x2+4xy+14y2 = 5 will not have terms in xy?
Solution :
Let the axes be rotated through an angle . Thus equation becomes

64. Solution Cont.
Therefore, the required angle is

65. Thank you