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
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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 )
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
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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)
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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
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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).
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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