Rectangular Cartesian Co-ordinates PDF

2 Rectangular Cartesian Co-ordinates 2 60 1.1 Introduction. E3 Co-ordinates of a point are the real variables associated in an order to a point to describe its location in some space. Here the space is the two dimensional plane. The work of describing the position of a Y point in a plane by an order...

2 Rectangular Cartesian Co-ordinates 2 60 1.1 Introduction. E3 Co-ordinates of a point are the real variables associated in an order to a point to describe its location in some space. Here the space is the two dimensional plane. The work of describing the position of a Y point in a plane by an ordered pair of real numbers can be done in different ways. Quadrant I Quadrant II The two lines XOX' and YOY' divide the plane in four quadrants. XOY, YOX', X' (+,+) (–,+) OY', Y'OX are respectively called the first, the second, the third and the fourth X' X quadrants. We assume the directions of OX, OY as positive while the directions of Quadrant III O Quadrant IV (+,–) (–,–) OX', OY' as negative. First quadrant Second quadrant Third quadrant Fourth quadrant + – – + ID x-coordinate y-coordinate point + + – – (+,+) (–,+) (–,–) (+,–) D YG U Quadrant Y' 1.2 Cartesian Co-ordinates of a Point. This is the most popular co-ordinate system. Let us consider two intersecting lines XOX' and YOY', which are perpendicular to each other. Let P be any point in the plane of lines. Draw the rectangle OLPM with its adjacent sides OL,OM along the Y lines XOX', YOY' respectively. The position of the point P can be fixed in the plane x P(x, y) provided the locations as well as the magnitudes of OL, OM are known. M y Axis of x : The line XOX' is called axis of x. X U Axis of y : The line YOY' is called axis of y. Co-ordinate axes : x axis and y axis together are called axis of ordinates or axis of reference. O L X coY ST Origin : The point ‘O’ is called the origin of co-ordinates or the origin. Oblique axes : If both the axes are not perpendicular then they are called as oblique axes. Let OL = x and OM = y which are respectively called the abscissa (or x-coordinate) and the ordinate (or ycoordinate). The co-ordinate of P are (x, y). Note :  Co-ordinates of the origin is (0, 0).  The y co-ordinate of every point on x-axis is zero.  The x co-ordinate of every point on y-axis is zero. 1.3 Polar Co-ordinates. Rectangular Cartesian Co-ordinates 3 Let OX be any fixed line which is usually called the initial line and O be a fixed point on it. If distance of any point P from the O is 'r' and XOP   , then (r,  ) are called the polar co-ordinates of a point P. Y P(r,) If (x, y) are the cartesian co-ordinates of a point P, then r x  r cos  ; y  r sin  and r  x 2  y 2  X 1.4 Distance Formula. Y' The distance between two points P(x 1 , y 1 ) and Q( x 2 , y 2 ) is given by :  The distance of a point M (x 0 , y 0 ) from origin O (0, 0 ) OM  (x 02  y 02 ). Q Y E3 PQ  (PR)2  (QR)2  (x 2  x 1 )2  (y 2  y1 )2 Note X O 60 y   tan 1   x X P R X O Y ID  If distance between two points is given then use  sign.  When the line PQ is parallel to the y-axis, the abscissa of point P and Q will be equal i.e, x 1  x 2 ; U  PQ | y 2  y 1 |  When the segment PQ is parallel to the x-axis, the ordinate of the points P and Q will be equal i.e., y 1  y 2. Therefore PQ | x 2  x 1 | D YG (1) Distance between two points in polar co-ordinates : Let O be the pole and OX be the initial line. Let P and Q be two given points whose polar co-ordinates are (r1 ,  1 ) and (r2 ,  2 ) respectively. Then OP  r1 , OQ  r2 P(r1,1) POX   1 and QOX   2 then POQ  (1   2 )  M If the point (x, y) be equidistant from the points (a  b, b  a) and (a  b, a  b) , then ax  by  0 (b) ax  by  0 (c) bx  ay  0 [MP PET 1983, 94] (d) bx  ay  0 Let points P(x , y ) , A (a  b, b  a), B(a  b, a  b). According to Question, PA  PB , i.e., PA 2  PB 2  (a  b  x )2  (b  a  y)2  (a  b  x )2  (a  b  y)2  (a  b)2  x 2  2 x (a  b)  (b  a)2  y 2  2 y(b  a) = (a  b)2  x 2  2 x (a  b)  (a  b)2  y 2  2y(a  b)  2 x (a  b  a  b)  2 y(b  a  a  b)   4 bx   4 ay  bx  ay  0 Example: 2 X :  Always taking  1 and  2 in radians. (a) Solution: (d) r2 2 PQ  r12  r22  2r1r2 cos(1   2 ) Note Example: 1 (1-2) 1 O (PQ)2  r12  r22  2r1r2 cos(1   2 ) ST  (OP ) 2  (OQ ) 2  (PQ) 2 2OP. OQ U In POQ, from cosine rule cos( 1   2 )  Q(r2,2) r1 If cartesian co-ordinates of any point are ( 3 ,1) , then its polar co-ordinates is 4 Rectangular Cartesian Co-ordinates 4 (a) (2,  / 3) (b) ( 2 ,  / 6) (c) (2,  / 6 ) (d) None of these We know that x  r cos  , y  r sin  Solution: (c)  3  r cos  , 1  r sin   1    /6 r  ( 3 ) 2  (1) 2  2 ,   tan 1    3 60 Polar co-ordinates  (2,  / 6 ). 1.5 Geometrical Conditions. (1) Properties of triangles E3 (i) In any triangle ABC , AB  BC  AC and | AB  BC |  AC. (ii) The ABC is equilateral  AB  BC  CA. (iii) The ABC is a right angled triangle  AB 2  AC 2  BC 2 or AC 2  AB 2  BC 2 or BC 2  AB 2  AC 2. (2) Properties of quadrilaterals ID (iv) The ABC is isosceles  AB  BC or BC  CA or AB  AC. D (i) The quadrilateral ABCD is a parallelogram if and only if (a) AB  DC, AD  BC , or (b) the middle points of BD and AC are the same,  U In a parallelogram diagonals AC and BD are not equal and   2 C B  B A B. D YG (ii) The quadrilateral ABCD is a rectangle if and only if (a) AB  CD, AD  BC and AC 2  AB 2  BC 2 or, (b) AB  CD, AD  BC , AC  BD or, (c) the middle points of AC and BD are the same and AC=BD. (    / 2 ) C D  B A ST U (iii) The quadrilateral ABCD is a rhombus (but not a square) if and only if (a) AB  BC  CD  DA and AC  BD or, (b) the middle points of AC and BD are the same and AB  AD but AC  BD. (   / 2) D C B  A B B (iv) The quadrilateral ABCD is a square if and only if (a) AB  BC  CD  DA and AC  BD or (b) the middle points of AC and BD are the same and AC  BD, (   / 2) , AB  AD. D Note C : Diagonals of square, rhombus, rectangle and parallelogram always bisect each other.  Diagonals of rhombus and square bisect each other at right angle.  Four given points are collinear, if area of quadrilateral is zero. Example: 3  A B ABC is an isosceles triangle. If the co-ordinates of the base are B(1,3) and C (– 2,7) the co-ordinates of vertex A can be [Orissa JEE 2002] Rectangular Cartesian Co-ordinates 5 (a) (1, 6) Solution: (c)  1  (b)   , 5   2  (c) 5   ,6 6  (d) None of these Let the vertex of triangle be A(x, y). Then the vertex A(x , y ) is equidistant from B and C because ABC is an isosceles triangle, therefore (x  1) 2  (y  3) 2 = (x  2) 2  (y  7) 2  6 x  8 y  43  0 5   ,6 6  60  1  Thus, any point lying on this line can be the vertex A except the mid point   , 5  of BC. Hence vertex A is  2  The extremities of diagonal of parallelogram are the points (3, – 4) and (– 6,5) if third vertex is (– 2,1), then fourth vertex is Example: 4 [Rajasthan PET 1987] (a) (1,0) (b) (–1,0) (c) (1,1) diagonal AC and BD coincide. So, x  2 6  3 y 1 5  4   and 2 2 2 2 x  1, y  0.  Coordinates of D are (–1, 0) ID Solution: (c) The vertices A and D of square ABCD lie on positive side of x and y-axis respectively. If the vertex C is the point (12, 17), then the coordinate of vertex B are Y (a) (14, 16) (b) (15, 3) M 12 C (c) (17, 5) (d) (17, 12) P Y  a 5 Let the co-ordinate of B be (h, k) a Draw BL and CM perpendicular to x-axis and y-axis. D  a cos   CM  OD  AL  12 and a sin   DM  OA  BL  5 U Example: 5 (d) None of these Let A(3,4 ) and C(6, 5) be the ends of diagonal of parallelogram ABCD. Let B(2,1) and D be (x, y), then mid points of E3 Solution: (b) 12  O  h  OL  OA  AL  5  12  17 A 5 12 Hence, Point B is (17, 5). A triangle with vertices (4, 0); (–1, –1); (3, 5) is Example: 6 Solution: (a) (a) Isosceles and right angled (b) Isosceles but not right angled (c) Right angled but not isosceles(d) Neither right angled nor isosceles Let A (4,0); B(1,1); C(3,5) then U So triangle is isosceles and also (BC )2  ( AB ) 2  ( AC )2. Hence  ABC is right angled isosceles triangle. ST 1.6 Section Formulae. If P( x , y ) divides the join of A( x 1 , y 1 ) and B(x 2 , y 2 ) in the ratio m 1 : m 2 (m 1 , m 2  0) (1) Internal division : If P( x , y ) divides the segment AB internally in the ratio of m 1 : m 2 PA m1  PB m 2 (x2,y2) B The co-ordinates of P( x , y ) are m x  m 2 x1 m y  m 2 y1 and y  1 2 x 1 2 m1  m 2 m1  m 2 5 L X [AIEEE 2002] AB  26 , AC  26 , BC  52 ; i.e. AB  AC  B a   k  BL  DM  OM  OD  17  12  5 D YG a P (x, y) (x1,y1) A (2) External division : If P( x , y ) divides the segment AB externally in the ratio of m 1 : m 2 6 Rectangular Cartesian Co-ordinates 6 PA m 1   PB m 2 P(x, y) (x2, y2) m x  m 2 x1 m y  m 2 y1 The co-ordinates of P( x , y ) are x  1 2 and y  1 2 m1  m 2 m1  m 2 :  If P( x , y ) divides the join of x A A( x 1 , y 1 ) and B(x 2 , y 2 ) in the ratio  : 1(  0) , then 60 Note B (x1,y1) x 2  x1 y  y1. Positive sign is taken for internal division and negative sign is taken for ;y 2  1  1 external division. E3  x 2 y1  y 2  ,  [Here m 1 : m 2 :: 1 : 1 ] 2 2   : 1. If  is positive, then divides internally and if  is negative, then  0 divides the join of points ID x  The mid point of AB is  1   For finding ratio, use ratio divides externally.  Straight line ax  by  c A( x 1 , y 1 ) and B(x 2 , y 2 ) in the U  ax  by 1  c  . ratio   1  ax 2  by 2  c  If ratio is –ve then divides externally and if ratio is +ve then divides internally. D YG The co-ordinate of the point dividing internally the line joining the points (4, –2) and (8, 6) in the ratio 7: 5 will be Example: 7 (a) (16, 18) Solution: (c)  19 8   ,   3 3 (c)  8 19  (d)  ,  3 3  Let point (x, y) divides the line internally. Then x  m y  m 2 y1 7(6 )  5(2) 8 m1 x 2  m 2 x 1 7(8 )  5(4 ) 19  . = , y 1 2 = 12 3 12 3 m1  m 2 m1  m 2 The line x  y  4 divides the line joining the points (–1,1) and (5, 7) in the ratio Example: 8 (a) 2 : 1 (b) 1 : 2 Internally (c) 1 : 2 Externally  ax  by 1  c   1  1  4  4 1  = – Required ratio =   1     5 7 4  8 2  ax 2  by 2  c  The line joining points (2, –3) and (–5, 6)is divided by y-axis in the ratio (a) 2 : 5 (b) 2 : 3 (c) 3 : 5 ST Solution: (a) [IIT 1965, UPSEAT 1999] (d) None of these (Internally) U Solution: (b) Example: 9 (b) (18, 16) [AMU 1979; MP PET 1984] Let ratio be k : 1 and coordinate of y-axis are (0, b). Therefore, 0  [MP PET 1999] (d) 1 : 2 k (5)  1(2) 2 k  k 1 5 1.7 Some points of a Triangle. (1) Centroid of a triangle : The centroid of a triangle is the point of intersection of its medians. The centroid divides the medians in the ratio 2 : 1 (Vertex : base) If A( x 1 , y 1 ) , B(x 2 , y 2 ) and C(x 3 , y 3 ) are the vertices of a triangle. If G be the centroid upon one of the median (say) AD, then AG : GD = 2 : 1  x  x 2  x 3 y1  y 2  y 3   Co-ordinate of G are  1 ,  3 3   Example: 10 A(x1,y1) 2 F B (x2,y2) The centroid of a triangle is (2,7)and two of its vertices are (4, 8) and (–2, 6) the third vertex is (a) (0, 0) (b) (4, 7) (c) (7, 4) (d) (7, 7) 1 2 1 G 1 D E 2 C (x3,y3) [Kerala (Engg.) 2002] Rectangular Cartesian Co-ordinates 7 Solution: (b) Let the third vertex (x, y) x 42 y86 , i.e. x  4 , y  7 2 ,7  3 3 Hence third vertex is (4, 7). 60 (2) Circumcentre : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. It is the centre of the circle which passes through the vertices of the A(x1,y1) triangle and so its distance from the vertices of the triangle is the same and this distance is known as the circum-radius of the triangle. E F Let vertices A, B, C of the triangle ABC be (x 1 , y1 ), (x 2 , y 2 ) and ( x 3 , y 3 ) and let circumcentre be O(x, y) and then (x, y) can be found by solving O (OA)2  (OB)2  (OC)2 E3 (x2,y2)B C(x3,y3) D i.e., (x  x1 )2  (y  y1 )2  (x  x 2 )2  (y  y 2 )2  (x  x 3 )2  (y  y 3 )2 :  Example: 11 Solution: (b) ID If angles of triangle i.e., A, B, C and vertices of triangle A(x 1 , y1 ), B(x 2 , y 2 ) and C (x 3 , y 3 ) are given, then circumcentre of the triangle ABC is  x 1 sin 2 A  x 2 sin 2 B  x 3 sin 2C y 1 sin 2 A  y 2 sin 2 B  y 3 sin 2C  ,   sin 2 A  sin 2 B  sin 2C sin 2 A  sin 2 B  sin 2C   U  If a triangle is right angle, then its circumcentre is the mid point of hypotenuse. If the vertices of a triangle be (2, 1); (5, 2) and (3,4) then its circumcentre is  13 9   13 9   9 13  (a)  ,  (b)  ,  (c)  ,   2 2  4 4 4 4  D YG Note [IIT 1964] (d) None of these Let circumcentre be O (x , y) and given points are A(2,1); B (5, 2); C (3, 4) and OA 2  OB 2  OC 2  (x  2) 2  (y  1) 2  (x  5) 2  (y  2) 2 and (x  2)  (y  1)  (x  3)  (y  4 ) 2 2 2 2.....(i).....(ii) 13 9 ,y  On solving (i) and (ii), we get x  4 4 A ST U (3) Incentre : The incentre of a triangle is the point of intersection of internal bisector of the angles. Also it is a centre of a circle touching all the sides of a triangle.  ax  bx 2  cx 3 ay1  by 2  cy3  Co-ordinates of incentre  1 ,  ab c ab c   Where a, b, c are the sides of triangle ABC. F B E I C D (4) Excircle : A circle touches one side outside the triangle and other two extended sides then circle is known as excircle. Let ABC be a triangle then there are three excircles with three excentres. Let I1 , I 2 , I 3 opposite to vertices A,B and C respectively. If vertices of triangle are A I2 I3 A(x 1 , y1 ), B(x 2 , y 2 ) and C ( x 3 , y 3 ) then   ax 1  bx 2  cx 3  ay1  by 2  cy 3 I1   , abc abc   ax  bx 2  cx 3 ay1  by 2  cy 3 I 2   1 , ab c ab c  Note    B C I1   ax  bx 2  cx 3 ay1  by 2  cy3   , I3   1 ,  ab c ab c    :  Angle bisector divides the opposite sides in the ratio of remaining sides e.g. BD AB c   DC AC b 8 Rectangular Cartesian Co-ordinates 8  Incentre divides the angle bisectors in the ratio (b  c) : a, (c  a) : b and (a  b) : c  Excentre : Point of intersection of one internal angle bisector and other two external angle bisector is called as excentre. There are three excentres in a triangle. Co-ordinate of each can be obtained by changing the sign of a,b,c respectively in the formula of in-centre. The incentre of the triangle with vertices (1, 3 ), (0, 0) and (2, 0) is Solution: (d)    1, 3   2    2 1   (b)  ,  3 3  (c)  Here AB  BC  CA  The triangle is equilateral. So, the incentre is the same as the centroid. 2 3   ,  3 2     1   (d)  1, 3   60 (a) [IIT Screening 2000] (1, 3) A 10 2 3 0 0   1     1, . ,  Incentre =     3 3 3    E3 Example: 12 2 2 ID 60o B (0, 0) C (2, 0) 2 then OE  BC , OF  AC , OD  AB U (5) Orthocentre : It is the point of intersection of perpendiculars drawn from vertices on opposite sides (called altitudes) of a triangle and can be obtained by solving the equation of any two A (x1, y1) altitudes. Here O is the orthocentre since AE  BC , BF  AC and CD  AB D F O Note D YG Solving any two we can get coordinate of O. B (x2, y2) C (x3, y3) E :  If a triangle is right angled triangle, then orthocentre is the ST U point where right angle is formed.  If the triangle is equilateral then centroid, incentre, orthocentre, circum-centre coincides.  Orthocentre, centroid and circum-centre are always collinear and centroid divides the line joining orthocentre and circum-centre in the ratio 2 : 1  In an isosceles triangle centroid, orthocentre, incentre, circum-centre lie on the same line. Example: 13 Solution: (c) The vertices of triangle are (0, 3) (– 3, 0) and (3, 0). The co-ordinate of its orthocentre are (a) (0, – 2) (b) (0, 2) (c) (0, 3) (d) (0, –3) Here AB  BC. In a right angled triangle, orthocentre is the point where right angle is formed.  Orthocentre is (0, 3) B (0,3) C (– 3, 0) Example: 14 O If the centroid and circumcentre of triangle are (3, 3); (6, 2), then the orthocentre is (a) (9, 5) (b) (3, –1) (c) (– 3, 1) [AMU 1991; DCE 1994] A (3, 0) [DCE 2000] (d) (– 3, 5) Rectangular Cartesian Co-ordinates 9 Let orthocentre be ( ,  ). We know that centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1 Solution: (d) 2(2)  1( ) 2(6 )  1( )    3 , 3    5 2 1 2 1 3  Hence orthocentre is (–3, 5). 1.8 Area of some Geometrical figures. 60 (1) Area of a triangle : The area of a triangle ABC with vertices A(x 1 , y1 ); B(x 2 , y 2 ) and C ( x 3 , y 3 ). The area of triangle ABC is denoted by ‘’and is given as y1 1 1 1  (x1 (y 2  y 3 )  x 2 (y 3  y1 )  x 3 (y1  y 2 ) 2 1 y2 y3 In equilateral triangle B (x2, y2) 3 2 a. 4 (ii) Having length of perpendicular as 'p' area is C (x3, y3) ID (i) Having sides a, area is E3 x1 1   x2 2 x3 A (x1, y1) (p 2 ). Note U 3 If a triangle has polar co-ordinates (r1 , 1 ), (r2 ,  2 ) and (r3 ,  3 ) then its area D YG :  1 [r1r2 sin( 2  1 )  r2r3 sin( 3   2 )  r3 r1 sin(1   3 )] 2  If area is a rational number. Then the triangle cannot be equilateral. (2) Collinear points : Three points A(x 1 , y 1 ); B(x 2 , y 2 ); C(x 3 , y 3 ) are collinear. If area of triangle is zero, x1 1 x2 (i)   0  2 x3 U i.e., y1 1 x1 y2 1  0  x 2 y3 1 x3 y1 1 y2 1  0 y3 1 (ii) AB  BC  AC or AC  BC  AB or AC  AB  BC ST (3) Area of a quadrilateral : If (x 1 , y1 ); (x 2 , y 2 ); (x 3 , y 3 ) and (x 4 , y 4 ) are vertices of a quadrilateral, then its 1 [(x 1 y 2  x 2 y 1 )  (x 2 y 3  x 3 y 2 )  (x 3 y 4  x 4 y 3 )  (x 4 y 1  x 1 y 4 )] 2 Area  Note : If two opposite vertex of rectangle are (x 1 , y1 ) and ( x 2 , y 2 ) , then its area is (y 2  y1 )(x 2  x1 ).  It two opposite vertex of a square are A(x 1 , y1 ) and C( x 2 , y 2 ) , then its area is  1 1 AC 2  [(x 2  x1 )2  (y 2  y1 )2 ] 2 2 (4) Area of polygon : The area of polygon whose vertices are (x 1 , y 1 );(x 2 , y 2 );(x 3 , y 3 );....( x n, y n ) is Example: 15 E3 60 10 Rectangular Cartesian Co-ordinates 10 1  | {( x 1 y 2  x 2 y 1 )  (x 2 y 3  x 3 y 2 ) ....  (x n y 1  x 1 y n )}| 2 Or Stair method : Repeat first co-ordinates one time in last for down arrow use positive sign and for up arrow use negative sign. x1 y1 x2 y2 x3 y3 1 1 : |  | {( x1y 2  x 2 y 3 ....  x n y1 )  (y1 x 2  y 2 x 3 ....  yn x1 )}|  Area of polygon = | : 2 2 : : xn yn x1 y1 The area of the triangle formed by the points (a, b  c), (b, c  a), (c, a  b) is [IIT 1963; EAMCET 1982; Rajasthan PET 2003] (b) a 2  b 2  c 2 (c) ab  bc  ca (d) 0 a ab c 1 a 1 1 a bc 1 1 ab c 1 b 1 1 =0 b c  a 1 = b b  c  a 1 , (Applying C2 C1  C2 ) = Area = 2 2 2 c cab 1 c 1 1 c ab 1 Example: 16 Three points are A(6, 3), B(– 3, 5), C(4, – 2) and P (x, y) is a point, then the ratio of area of PBC and ABC is x y2 7 (a) Example: 17 (a) (8, 8) Solution: (a) (c) x y2 7 [IIT 1983] (d) None of these 1  3(2  y)  4(y  5)  x (5  2) 7 x  7 y  14 x  y  2 Area of PBC  =  2 1 49 7 Area of ABC 6(5  2)  3(2  3)  4(3  5) 2 If the points (2K, K)(K, 2K) and (K, K) with K  0 enclose triangle of area 18 square units then the centroid of triangle is equal to D YG Solution: (a) (b) x y2 2 U Solution: (d) ID (a) abc  2K 1 K 2 K (b) (4, 4) (c) (– 4, – 4) (d) (4 2 , 4 2 ) K 1 K2 2 K 1  18   18  K   6. Consider K = +6 because K > 0, then the points (12, 6) (6,12) and (6,6). 2 K 1 U  12  6  6 6  12  6  , Hence, centroid =    (8 , 8 ) 3 3   2   1 , If the points (x+1, 2); (1,x + 2);  [Rajasthan PET 2002]  are collinear, then x is  x 1 x 1  (a) 4 (b) 0 (c) – 4 (d) None of these 2   1 , Let A  (x  1, 2) ; B  (1, x  2) ; C   . A, B, C are collinear, if area of ABC  0  x 1 x 1 ST Example: 18 Solution: (b,c)   Example: 19 x 1 1 1 x 1 x 2 1 x2 1 0  1 1 2 1 x 1 x 1 x 1 1 x 1 0 x 3 3 x 1 0 1 0 x 0 x2 1 0 2 1 x 1 (R1 R1  R2 ) (C2 C1  C2 )  x 2 (x  4 )  0  x  0,  4 1 The points (1, 1); (0, sec 2  ); (co sec 2 , 0) are collinear for (a)   n / 2 (b)   n / 2 [Roorkee 1963] (c)   n (d) None of these Rectangular Cartesian Co-ordinates 11 Solution: (b) The given points are collinear, if Area of   1 2 1 0 cosec 2 1 sec 2  0 1 1  0  1(sec 2  )  1(cosec 2 )  1(cosec 2. sec 2  )  0 1 1 1 1 1 1    0 00 0  2 2 2 2 2 2 2 cos  sin  sin . cos  sin . cos  sin . cos 2  n n Therefore the points are collinear for all value of  , except only   because at   , sec 2    (Not defined). 2 2 The points (0, 8/3) (1, 3) and (82, 30) are the vertices of [IIT 1983; Rajasthan PET 1988] (a) An equilateral triangle (b) An isosceles triangle (c) A right angled triangle (d) None of these Here A  (0, 8 / 3), B  (1,3) and C  (82, 30 ) Example: 20 Solution: (d) Since AB  BC  (1  81) 10 10 , AC  (82)2  (30  8 / 3)2  82 9 9 E3 AB  1  1 / 9  10 / 9 , BC  (81)2  (27 )2  27 10 = 81 60  10 10  82  AC.  Points A, B, C are collinear. 9 9 ID 1.9 Transformation of Axes. (1) Shifting of origin without rotation of axes : Let P  (x , y ) with respect to axes OX and OY. U Let O '  ( ,  ) with respect to axes OX and OY and let P  (x ' , y ' ) with respect to axes O'X' and O'Y', where OX and O'X' are parallel and OY and O'Y' are parallel. Y Then x  x ', y  y '   or x '  x   , y '  y   Y D YG Thus if origin is shifted to point ( ,  ) without rotation of axes, then new equation of curve can be obtained by putting x   in place of x and y   in place of y. P(x, y) (x', y') y (, ) X x O X O x (2) Rotation of axes without changing the origin : Let O be the origin. Let P  (x , y ) with respect to axes OX and OY and let P  (x ' , y ' ) with respect to axes OX and OY where Y Y' X ' OX  YOY '   x  x ' cos   y ' sin  then P (x,y) (x,y) y and U y  x ' sin   y ' cos  y  x '  x cos   y sin  X' x ST y '   x sin   y cos  x O  X X The above relation between ( x , y ) and (x ' , y ' ) can be easily obtained with the help of following table x y x' cos  sin  y'  sin  cos  Y (3) Change of origin and rotation of axes : If origin is changed to O ' ( ,  ) and axes are rotated about the new origin O ' by an angle  in the Y' P(x, y) (x', y')  X'  O O X 12 Rectangular Cartesian Co-ordinates 12 anticlock-wise sense such that the new co-ordinates of P(x , y ) become ( x ' , y ' ) then the equations of transformation 60 will be x    x ' cos   y ' sin  and y    x ' sin   y ' cos  (4) Reflection (Image of a point) : Let ( x , y ) be any point, then its image with respect to Example: 21 (ii) y-axis  ( x , y ) (iii) origin  ( x ,y ) (iv) line y  x  (y, x ) E3 (i) x axis  ( x , y ) The point (2,3) undergoes the following three transformation successively, (i) Reflection about the line y  x. (ii) Transformation through a distance 2 units along the positive direction of y-axis. (iii) Rotation through an angle of 45o about the origin in the anticlockwise direction. The final coordinates of points are Solution: (b)  1 7     2, 2    1 7   , (b)  2   2  1 7     2, 2   ID (a) (c) U     D YG Reflecting the point (2, –1) about y-axis, coordinate axes are rotated at 45 o angle in negative direction without shifting the origin. The new coordinates of the point are (a) Solution: (a) (d) None of these (i) The new position after reflection is (3,2) (ii) After transformation, it is (3, 2+ 2), i.e, (3, 4)  1 7 , (iii) Rotation makes it (3 cos 45 o  4 sin 45 o ,3 sin 45 o  4 cos 45 o ) , i.e.  2  2 Example: 22 [Roorkee 2000]  1 3    2, 2   3 1   , (b)  2   2 (c)  1 3   2, 2      (d) None of these The new position after reflection is (–2, –1)  1  3  , Rotation makes it [(2) cos(45 o )  (1) sin(45 o ),  (2) sin(45 o )  (1) cos(45 o )] , i.e.,   2  2 The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The co-ordinate of the point thus obtained are [DCE 1997] (a) (3, –3) (b) (–3, 3) (c) (3, 3) (d) (–3, –3) Reflection in the y-axis of the point (3,2) is (–3, 2) when it moves towards the negative side of y- axis through 5 units, then the new position is (–3, 2– 5) =(– 3, – 3) U Example: 23 ST Solution: (d) Y Q (–3, 2) P (3, 2) 2 X X O R 3 (– 3, – 3) Y 1.10 Locus. Locus : The curve described by a point which moves under given condition or conditions is called its locus. Equation to the locus of a point : The equation to the locus of a point is the relation, which is satisfied by the coordinates of every point on the locus of the point. Rectangular Cartesian Co-ordinates 13 Algorithm to find the locus of a point Step I : Assume the coordinates of the point say (h, k) whose locus is to be found. Step II : Write the given condition in mathematical form involving h , k. Step III : Eliminate the variable (s), if any. Step IV : Replace h by x and k by y in the result obtained in step III. The equation so obtained is the locus of the point which moves under some stated condition (s) :  Locus of a point P which is equidistant from the two point A and B is a straight line and is a 60 Note perpendicular bisector of line AB.  In above case if PA = kPB where k  1 , then the locus of P is a circle.  Locus of P if A and B is fixed. (a) Circle, if APB = constant  E3 (b) Circle with diameter AB , if APB  2 (c) Ellipse, if PA +PB = constant (d) Hyperbola, if PA – PB = constant Let A (2, – 3) and B( – 2, 1) be vertices of triangle ABC. If the centroid of this triangle moves on the line 2 x  3 y  1 , then the locus of the vertex C is the line [AIEEE 2004] (a) 3 x  2 y  3 (b) 2 x  3 y  7 (c) 3 x  2 y  5 (d) 2 x  3 y  9 Solution: (d) Let third vertex C be ( ,  ) ID Example: 24  2  2   3  1       2  ,  Centroid =    , i.e.  , 3 3  3  3  U      2  According to question, 2   3   1  2  3   6  3  2  3   9 3  3  Example: 25 The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is [IIT 1987; Rajasthan PET 1997] (a) Solution: (c) D YG Hence, locus of vertex C is 2 x  3 y  9. 36 x 2  9 y 2  4 l 2 36 x 2  9 y 2  l 2 (b) AP : PB  1 : 2 , then h  (c) 9 x 2  36 y 2  4 l 2 1  0  2  a 2a 3h  , Similarly b = 3k or a  12 3 2 (d) None of these Y B(0, b) 2  3h  Now we have OA  OB  AB     (3 k )2  l 2  2  2 2 2 2 b P(h, k) 1 U Hence locus of P (h, k) is given by 9 x 2  36 y 2  4 l 2 O a A (a,0) Example: 26 If A and B are two fixed points and P is a variable point such that PA  PB  4 , then the locus of P is a/an Solution: (b) (a) Parabola (b) Ellipse (c) Hyperbola We know that, PA  PB = constant. Then locus of P is an ellipse. ST X [IIT 1989; UPSEAT 2001] *** (d) None of these