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

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 )

8. Representation of points on plane

9. Co-ordinate of origin (0,0)

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

14. PROBLEMS ON COLLINEARITY OF THREE POINTS POINTS A , B and C ARE said to be collinear if AB +BC = AC

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

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

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)

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

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).

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