Tiêu đề XI - maths - chapter 10 - the straight lines-i.pdf ✅

Nishith Multimedia India (Pvt.) Ltd. 1 IX-Mathematics e-Techno Text Book CO-ORDINATE GEOMETRY Synopsis Date: 16-08-2010 Introduction: COORDINATE PLANE Let A,B be two points in a plane. Generally the distance between A and B denoted by AB.The line segment joining A and B is denoted by AB . The ray from A and passing through B denoted by AB  . The ray from A and passing through B is denoted by AB  . The line passing through A and B is denoted by AB  Let x x  and y y  be two mutually perpendicualr coordinate lines in a plane intersecting at O. The point O is called origin. The line x ox   is called x - axis as horizontal line and y- axis as vertical line. Let p be a point in the plane. Let L,M be the projections (feet of the perpendiculars ) of P on x - axis and y-axis respectively. Let x,y be the real numbers assigned to L,M on the lines x ox,y oy     respectively. y P x M(y) L(x) y x O Then x is called x - coordinate of P and y is called y - coordinate of P. The point P is denoted by (x,y). Every point in the plane can be represented by two numbers x,y (coordinates) and every pair of real numbers determine a point in the plane. x- axis and y-axis are called coordinate axes and the plane is called coordinate plane. The coordinate axes divide the plane into four equal parts. Each part is called a quadrant. The regions xoy, yo x , x oy   , y ox  are called I,II,III,IV quadrants respectively. Let P(x,y) be a point in the coordinate plane. Then 1) P lies in I quadrant iff x > 0, y > 0 2) P lies in II quadrant iff x < 0, y >0 3) P lies in III quadrant iff x <0, y <0 4) P lies in IV quadrant iff x > 0, y < 0 5) P lies in the x - axis iff y = 0. 6) P lies in the y - axis iff x = 0. 7) If P = O, then x = 0, y = 0 i.e., the coordinates of the origin O are (0,0) .
Nishith Multimedia India (Pvt.) Ltd. 2 IX-Mathematics e-Techno Text Book y x y x O II I III IV Note : LetA x ,y ,B x ,y  1 1 2 2    be two points in the coordinate plane. Then (i) AB  is a horizontal line iff y y 1 2  and (ii) AB  is a vertical line iff x x 1 2  . We state some results without proofs in this chapter which were learnt in previous classes. Distance between two points Let A  x y B x y 1 1 2 2 , , ,    be any two points on a line not parallel to the axes. From the adjacent figure we have the right angle triangle ABC. 2 2 2 AB AC BC   But AC x x ,BC y y     2 1 2 1     2 2 2      AB x x y y 2 1 2 1     2 2      AB x x y y 2 1 2 1 NOTE : The distance to the point A x , y  1 1  from origin is 2 2 1 1 x y  Illustration : 1. Find the distance between the points (1, 2) and (3, 2) Solution : Let A x ,y 1,2    1 1    B =x ,y 2 2  = 3,2
Nishith Multimedia India (Pvt.) Ltd. 3 IX-Mathematics e-Techno Text Book  Distance of     2 2 AB x x y y     2 1 2 1 =     2 2 3 1 2 2    =     2 2 2 0  = 2 units Illustration : 2. Find the distance to the point (2,1) from origin Solution : Let A= 1 1 (x ,y ) (2,1)   Distance of 2 2 OA x y  1 1 = 2 2 2 1  5 units Collinear points : The points which lie on the same line are called collinear points, If three points A, B, C are collinear then AB + BC = AC (or) AC + CB = AB (or) BA + AC = BC Note 1) : If A,B,C are collinear then A,B,C lie on the same line Note 2) : A,B,C are collinear , then the area of  ABC is zero Illustration : Show that the points (-1, 7), (3, -5), (4, -8) are collinear Solution : Let A = (-1, 7), B = (3, -5), C = (4, -8) are given points the distance between two points  x y x y 1 1 2 2 , , ,    is     2 2 x x y y 2 1 2 1     AB     2 2      3 1 5 7   2 2    4 12   16 144  160  4 10 units BC     2 2      4 3 8 5 2 2   1 3  10 units CA     2 2      1 4 7 8   25 225  250  5 10 units Now, AB+BC   4 10 10  5 10  AC  A B C , , are collinear.
Nishith Multimedia India (Pvt.) Ltd. 4 IX-Mathematics e-Techno Text Book Dividing a line segment in a given ratio (section formulae) : (P divides AB in the ratio m : n internally.) (P divides AB in the ratio m : n externally.) Section formulae : The point ‘P’ which divides the line segment joining the points A x y B x y  1 1 2 2 , , ,   in the ratio m:n i) internally is 2 1 2 1 , ; 0             mx nx my ny m n m n m n ii) externally is 2 1 2 1 , ;            mx nx my ny m n m n m n Illustration : Find the points which divide the line segment joining the points (1, –3), (–3, 9) in the ratio 1:2 internally and externally. Solution : Let A x y    ( , ) 1, 3 1 1  , B x y     2 2 , 3,9    and ratio =m :n = 1 : 2 The point which divides AB in the ratio 1:2 internally is 2 1 2 1 , mx nx my ny m n m n            1 3 2 1 1 9 2 3 , 1 2 1 2                  1,1 3         The point which divides AB in the ratio 1:2 externally is 2 1 2 1 , mx nx my ny m n m n            1 3 2 1 1 9 2 3 , 1 3 1 3                  3 2 1 9 6 5 15 , , 2 2 2 2                      Mid point of a line segment : Let A x y B x y    1 1 2 2 , , ,    If P is the mid point of the line segment AB then P divides AB in the ratio 1:1 internally. Let m:n=1:1