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co-ordinate geometry presented by Gurpreet kaur T.G.T. MATHS KV Faridkot
Rene Descartes A French mathematician who discovered the co-ordinate geometry. He was the first man who unified Algebra and Geometry. According to him every point of the plane can be represented uniquely by two numbers
co-ordinate geometry It is that branch of geometry in which two numbers called  co-ordinates  , are used to locate the position of a point in a plane . It is also called as Cartesian-geometry
Axes of reference ,[object Object],[object Object],[object Object],[object Object]
 
CO-ORDINATES OF A POINT ,[object Object],[object Object]
REPRESENTATION OF A POINT THE CO-ORDINATES OF A POINT IS ALWAYS REPRESENTED  BY ORDERED -PAIR (  ) FIRST PUT X-CO-ORDINATE THEN Y-CO-ORDINATE IN BRACKET ( X, Y )
Representation of points on plane
Co-ordinate of origin (0,0)
Distance formula To find out the distance between two points  in the plane
Let the  two points are P(x 1 ,y 1 ) and Q(x 2 ,y 2 )  P(x 1 ,y 1 ) Q(x 2 ,y 2 )
P(x 1 ,y 1 ) Q(x 2 ,y 2 ) x 1 x 2 y 2 y 1 R(x 2 ,y 1 ) Then  distance QR =  y 2  - y 1 and  PR =   x 2  - x 1
[object Object],[object Object],[object Object],[object Object],[object Object]
PROBLEMS   ON   COLLINEARITY   OF   THREE   POINTS POINTS A , B and C ARE said to be collinear  if  AB +BC = AC
PROBLEM 1 Determine by distance formula , whether the points (2,5) , (-1,2) and (4,7) are collinear. ,[object Object],[object Object],[object Object],[object Object]
SIMILARLY  BC=   (4+1)  2  +(7-2)  2 =   (5)  2  +(5)  2 =   25+25=   50 =   25X2= 5   2 AC=   (2-4)  2  +(5-7)  2 =   (-2)  2  +(-2)  2  =   4+4=   8 =2   2  THUS  3   2+ 2   2 =5   2      AB +AC = BC POINTS  B , A  C  are collinear
Problem on equidistant ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Note  ,[object Object]
Note   ,[object Object],[object Object]
For square ,[object Object],[object Object],For rhombus 1.all sides are equal but 2. diagonals are  not equal
For  rectangle 1. Opposite sides are equal 2. Diagonals are equal For  parallelogram 1. Opposite sides are equal 2. Diagonals bisect each other
section formula It gives the co-ordinates of the point  which divides the given line segment  in the ratio   m : n
Let AB  is a line segment  joining the points A(x 1 ,y 1 ) and B(x 2 ,y 2 )  LET  P(x,y) be a point  which divides line segment AB  in the ratio   m : n  internally   therefore ( x 1 ,y 1 ) ( x 2 ,y 2 )
(x 1 ,y 1 ) (x 2 ,y 2 ) (x,y 1 ) (x-x 1 ) (x 2 ,y) (x 2   -  x) (y 2   -  y) (y-y 1 ) x 1 x x 2 Complete  the figure as follows
Now   AQP        PRB …..by AA similarity   AP/PB = AQ/PR = PQ/BR    BY CPST  m/n  = x-x 1 /x 2 -x =y-y 1 /y 2 -y m/n  = x-x 1 /x 2 -x solving this equation  for x we will get x = mx 2 +nx 1 m+n Similarly solving the equation  m/n  =y-y 1 /y 2 -y  we will get y = my 2 +ny 1 m+n (x 1 ,y 1 ) (x-x 1 ) (y-y 1 ) (y 2   -  y) (x 2   -  x)
Thus if a point p(x,y) divides a line segment joining the points A(x 1 ,y 1 )  and  B(x 2 ,y 2 ) in the ratio  m : n  then the co-ordinates  of  P  are given by
[object Object],PROBLEM 1
SOLUTION 1 Let co-ordinates of point R (X,Y) X = 4[2]+3[1] 4+3 X= {8+3]/7 = 11/7 Here  m=4 , n=3,  x 1 =1 , y 1 =2  x 2 = 2 ,  y 2  = 3 Y = 4[3]+3[2] 4+3 Y = [ 12+6]/7 = 18/7 Hence co-ordinates of R = (11/7 , 18/7)
[object Object],PROBLEM 2
Let  k : 1 be the required ratio   Here x=11 , y=15 , x 1 = 15 , y 1 =5  ,x 2 =9 , y 2 =20 m= k  ,  n= 1 Therefore  using section formula 11 = 9k+15 k+1 and 15  =  20k+5 k+1 solve either of these two equation let’s take first equation solution 2 11k+11=9k+15 11k-9k=15-11 2k=4 k=2 So,the required ratio is 2:1
[object Object],Problem 3
Solution 3 Let k:1 is the required ratio. P(x,0) point on x-axis which divides AB in required ratio. Here x=x , y=0 , x 1 = 6 , y 1 =4  ,x 2 =1 , y 2 =  -7 Therefore  using section formula x = k+6 k+1 and 0= -7k+4 k+1 solve either of these two equation let’s take second equation 0= -7k+4 7k=4 K= 4/7 So,the required ratio is 4:7.
AREA OF TRIANGLE   Let vertices of triangle are  ( x 1 ,y 1 ) (x 2 ,y 2 )  and  (x 3 ,y 3 ). Area of Triangle  =1/2 ( x1 ( y2-y3 ) +x2 ( y3-y1 ) +x3 ( y1-y2 ) ) A( x1,y1) B(x2,y2) C(x3,y3).
PROBLEM 2 ,[object Object]
solution 2 Here x 1 = 6 , y 1 =4  ,x 2 =1 , y 2 = -7, x 3 =2 , y 3 =3 Area of Triangle  =1/2 ( x1 ( y2-y3 ) +x2 ( y3-y1 ) +x3 ( y1-y2 ) ) Area of Triangle  =1/2 ( 6 ( -7-3 ) +1 ( 3-4 ) +2 (4 +7 ) ) =1/2(-60-1+22) =1/2(-39) = -39/2 = -39/2 = 39/2 So,Area of Triangle is 19.5

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Coordinategeometry1 1

  • 1. co-ordinate geometry presented by Gurpreet kaur T.G.T. MATHS KV Faridkot
  • 2. Rene Descartes A French mathematician who discovered the co-ordinate geometry. He was the first man who unified Algebra and Geometry. According to him every point of the plane can be represented uniquely by two numbers
  • 3. co-ordinate geometry It is that branch of geometry in which two numbers called co-ordinates , are used to locate the position of a point in a plane . It is also called as Cartesian-geometry
  • 4.
  • 5.  
  • 6.
  • 7. REPRESENTATION OF A POINT THE CO-ORDINATES OF A POINT IS ALWAYS REPRESENTED BY ORDERED -PAIR ( ) FIRST PUT X-CO-ORDINATE THEN Y-CO-ORDINATE IN BRACKET ( X, Y )
  • 10. Distance formula To find out the distance between two points in the plane
  • 11. Let the two points are P(x 1 ,y 1 ) and Q(x 2 ,y 2 ) P(x 1 ,y 1 ) Q(x 2 ,y 2 )
  • 12. P(x 1 ,y 1 ) Q(x 2 ,y 2 ) x 1 x 2 y 2 y 1 R(x 2 ,y 1 ) Then distance QR = y 2 - y 1 and PR = x 2 - x 1
  • 13.
  • 14. PROBLEMS ON COLLINEARITY OF THREE POINTS POINTS A , B and C ARE said to be collinear if AB +BC = AC
  • 15.
  • 16. SIMILARLY BC=  (4+1) 2 +(7-2) 2 =  (5) 2 +(5) 2 =  25+25=  50 =  25X2= 5  2 AC=  (2-4) 2 +(5-7) 2 =  (-2) 2 +(-2) 2 =  4+4=  8 =2  2 THUS 3  2+ 2  2 =5  2  AB +AC = BC POINTS B , A C are collinear
  • 17.
  • 18.
  • 19.
  • 20.
  • 21.
  • 22. For rectangle 1. Opposite sides are equal 2. Diagonals are equal For parallelogram 1. Opposite sides are equal 2. Diagonals bisect each other
  • 23. section formula It gives the co-ordinates of the point which divides the given line segment in the ratio m : n
  • 24. Let AB is a line segment joining the points A(x 1 ,y 1 ) and B(x 2 ,y 2 ) LET P(x,y) be a point which divides line segment AB in the ratio m : n internally therefore ( x 1 ,y 1 ) ( x 2 ,y 2 )
  • 25. (x 1 ,y 1 ) (x 2 ,y 2 ) (x,y 1 ) (x-x 1 ) (x 2 ,y) (x 2 - x) (y 2 - y) (y-y 1 ) x 1 x x 2 Complete the figure as follows
  • 26. Now  AQP   PRB …..by AA similarity AP/PB = AQ/PR = PQ/BR  BY CPST m/n = x-x 1 /x 2 -x =y-y 1 /y 2 -y m/n = x-x 1 /x 2 -x solving this equation for x we will get x = mx 2 +nx 1 m+n Similarly solving the equation m/n =y-y 1 /y 2 -y we will get y = my 2 +ny 1 m+n (x 1 ,y 1 ) (x-x 1 ) (y-y 1 ) (y 2 - y) (x 2 - x)
  • 27. Thus if a point p(x,y) divides a line segment joining the points A(x 1 ,y 1 ) and B(x 2 ,y 2 ) in the ratio m : n then the co-ordinates of P are given by
  • 28.
  • 29. SOLUTION 1 Let co-ordinates of point R (X,Y) X = 4[2]+3[1] 4+3 X= {8+3]/7 = 11/7 Here m=4 , n=3, x 1 =1 , y 1 =2 x 2 = 2 , y 2 = 3 Y = 4[3]+3[2] 4+3 Y = [ 12+6]/7 = 18/7 Hence co-ordinates of R = (11/7 , 18/7)
  • 30.
  • 31. Let k : 1 be the required ratio Here x=11 , y=15 , x 1 = 15 , y 1 =5 ,x 2 =9 , y 2 =20 m= k , n= 1 Therefore using section formula 11 = 9k+15 k+1 and 15 = 20k+5 k+1 solve either of these two equation let’s take first equation solution 2 11k+11=9k+15 11k-9k=15-11 2k=4 k=2 So,the required ratio is 2:1
  • 32.
  • 33. Solution 3 Let k:1 is the required ratio. P(x,0) point on x-axis which divides AB in required ratio. Here x=x , y=0 , x 1 = 6 , y 1 =4 ,x 2 =1 , y 2 = -7 Therefore using section formula x = k+6 k+1 and 0= -7k+4 k+1 solve either of these two equation let’s take second equation 0= -7k+4 7k=4 K= 4/7 So,the required ratio is 4:7.
  • 34. AREA OF TRIANGLE Let vertices of triangle are ( x 1 ,y 1 ) (x 2 ,y 2 ) and (x 3 ,y 3 ). Area of Triangle =1/2 ( x1 ( y2-y3 ) +x2 ( y3-y1 ) +x3 ( y1-y2 ) ) A( x1,y1) B(x2,y2) C(x3,y3).
  • 35.
  • 36. solution 2 Here x 1 = 6 , y 1 =4 ,x 2 =1 , y 2 = -7, x 3 =2 , y 3 =3 Area of Triangle =1/2 ( x1 ( y2-y3 ) +x2 ( y3-y1 ) +x3 ( y1-y2 ) ) Area of Triangle =1/2 ( 6 ( -7-3 ) +1 ( 3-4 ) +2 (4 +7 ) ) =1/2(-60-1+22) =1/2(-39) = -39/2 = -39/2 = 39/2 So,Area of Triangle is 19.5