Co-Ordinate Geometry Study Guide

CO-ORDINATE GEOMETRY Draw PM and PN perpendiculars on OX and OY respectively. OM (or MP) is called the y co-ordinate or the ordinate of the point P. CONTENTS  Definition  Cartesian Co-ordinates  Distance Between Two Points  Section Formula  Application of Section Formula  Area of Triangles Axes of Co-ordinates In the figure OX and OY are called as x-axis and y-axis respectively and both together are known as axes of co-ordinates. Origin It is point O of intersection of the axes of co-ordinates. Co-ordinates of the Origin It has zero distance from both the axes so that its abscissa and ordinate are both zero. Therefore, the coordinates of origin are (0, 0). Abscissa CO-ORDINATE GEOMETRY It is a branch of geometry which sets up a definite correspondence between the position of a point in a plane and a pair of algebraic numbers, called co-ordinates.  CARTESIAN CO-ORDINATES (RECTANGULAR CO-ORDINATES) The distance of the point P from y-axis is called its abscissa. In the figure, OM is the Abscissa. Ordinate The distance of the point P from x-axis is called its ordinate. ON is the ordinate in the figure. Quadrant    In Cartesian co-ordinates the position of a point P is determined by knowing the distances from two perpendicular lines passing through the fixed point. Let O be the fixed point called the origin and XOX and YOY, the two perpendicular lines through O, called Cartesian or Rectangular co-ordinates axes. The axes divide the plane into four parts. These four parts are called quadrants. So, the plane consists of axes and quadrants. The plane is called the cartesian plane or the coordinate plane or the xy-plane. These axes are called the co-ordinate axes. A quadrant is co-ordinate axes. 1 4 part of a plane divided by Y Y N y X' M O Y' II P(x,y) x O X' X I III X IV Y' (i) XOY is called the first quadrant (ii) YOX the second. (ii) Ordinate = PM = ON = 4 units (iii) XOY the third. (iii) Co-ordinates of the point P = (Abscissa, ordinate) = (3, 4) (iv) YOX the fourth as marked in the figure. Ex.2  RULES OF SIGNS OF CO-ORDINATES   Determine (i) Abscissa (ii) ordinate (iii) Co-ordinates of point P given in the following figure. Y (i) In the first quadrant, both co-ordiantes i.e., abscissa and ordinate of a point are positive. P (ii) In the second quadrant, for a point, abscissa is negative and ordinate is positive. X' M (iii) In the third quadrant, for a point, both abscissa and ordinate are negative. (iv) In the fourth quadrant, for a point, the abscissa is positive and the ordinate is negative. Sol. (i) Abscissa of the point P = – NP = – OM = – a (ii) Ordinate of the point P = MP = ON = b = (–a, b) Ex.3 X Y' Write down the (i) abscissa (ii) ordinate (iii) Co-ordinates of P, Q, R and S as given in the figure. x-co-ordinate y-co-ordinate Point First quadrant + + (+,+) Second quadrant – + (–,+) Third quadrant – – (–,–) Fourth quadrant + – (+,–) 2 3 X' Sol. X O 3 5 1 5 S Y' Point P Abscissa of P = 2; Ordinate of P = 3 Y N Co-ordinates of P = (2, 3) P Point Q 4 O P 4 EXAMPLES  From the adjoining figure find Y 2 Q R X' X O (iii) Co-ordinates of point P = (abscissa, ordinate) I II (–,+) (+,+) X' III O IV (–,–) (+,–) Ex.1 a b Y' Y Quadrant N Abscissa of Q = – 2; Ordinate of Q = 4 3 M X Y' (i) Abscissa (ii) Ordinate (iii) Co-ordinates of a point P Sol. (i) Abscissa = PN = OM = 3 units Co-ordinate of Q = (–2, 4) Point R Abscissa of R = – 5; Ordinate of R = – 3 Co-ordinates of R = (–5, –3) Point S Abscissa of S = 5; Ordinate of S = – 1 Co-ordinates of S = (5, – 1) Ex.4 Draw a triangle ABC where vertices A, B and C are (0, 2), (2, – 2), and (–2, 2) respectively. Sol. Plot the point A by taking its abscissa O and ordinate = 2. Y A(4, 6) B(–2, 3) Similarly, plot points B and C taking abscissa 2 and –2 and ordinates – 2 and 2 respectively. Join A, B and C. This is the required triangle. X' C(–2, –5) Y D(4, –7) C(–2, 2) A(0, 2) X' Sol. Y'  DISTANCE BETWEEN TWO POINTS X O B(2, –2) Draw a rectangle PQRS in which vertices P, Q, R and S are (1, 4), (–5, 4), (–5, –3) and (1, – 3) respectively. Plot the point P by taking its abscissa 1 and ordinate – 4. P(x1, y1) and Q (x2, y2) is given by PQ = Join the points PQR and S. PQRS is the required rectangle. Y Q(–5, 4) ( x 2  x1 ) 2  ( y 2  y1 ) 2 i.e., PQ = (Diff . of abscissa) 2  (Diff . of ordinates) 2 Note : If O is the origin and P(x, y) is any point, then from the above formula, we have Similarly, plot the points Q, R and S taking abscissa as –5, –5 and 1 and ordinates as 4, – 3 and –3 respectively. OP = ( x  0) 2  ( y  0) 2 = x 2  y2 EXAMPLES  Ex.7 Find the distance between the points (i) P(–6, 7) and Q(–1, –5) (ii) R(a + b, a – b) and S(a – b, –a – b) (iii) A(at12 ,2at1 ) and B(at 22 ,2at 2 ) P(1, 4) Sol. (i) Here, X' X O R(–5, –3) S(1, –3) Y' Ex.6 Sol.    Theorem : The distance between two points Y' Ex.5 X O Draw a trapezium ABCD in which vertices A, B, C and D are (4, 6), (–2, 3), (–2, –5) and (4, –7) respectively. Plot the point A taking its abscissa as 4 and ordinate as 6. Similarly plot the point B, C and D taking abscissa as – 2, –2 and 4 and ordinates as 3, – 5, and –7 respectively. Join A, B, C and D ABCD is the required trapezium. x1 = – 6, y1 = 7 and x2 = – 1, y2 = – 5 ( x 2  x1 ) 2  ( y 2  y1 ) 2 PQ =  PQ = ( 1  6) 2  (5  7) 2  PQ = 25  144 = 169 = 13 (ii) We have, RS = (a  b  a  b) 2  (a  b  a  b) 2 4b 2  4a 2 = 2 a 2  b 2  RS = (iii) We have, AB = (at 2 2  at12 ) 2  (2at 2  2at1 ) 2 AB = a 2 ( t 2  t1 ) 2 ( t 2  t1 ) 2  4a 2 ( t 2  t1 ) 2 AB = a(t2 – t1) AB = (a  a ) 2  (a  a ) 2 = 4a 2  4a 2 =2 2 a ( t 2  t1 ) 2  4 BC = Ex.8 If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay. Sol. Let P(x, y), Q(a + b, b – a) and R (a – b, a + b) be the given points. Then PQ = PR  BC = =a 8 =2 2 a (Given) {x  (a  b)}2  {y  (a  b)}2 and, AC = ( 3a  a 2 )  ( 3a  a ) 2  AC = a 2 ( 3  1)  a 2 ( 3  1) 2  AC = a 3  1  2 3  3  1  2 3  {x – (a + b)}2 + {y – (b – a)}2 =a 8 =2 2a = {x – (a – b)}2 + {y – (a + b)}2 Clearly, we have  x2 – 2x (a + b) + (a + b)2 AB = BC = AC + y2 – 2y (b – a) + (b – a)2 Hence, the triangle ABC formed by the given points is an equilateral triangle. = x2 + (a – b)2 – 2x(a – b) + y2 – 2 (a + b) + (a + b)2  a 2 (1  3 ) 2  a 2 ( 3  1) 2  BC = a 1  3  2 3  1  3  2 3  {x  (a  b)}2  {y  ( b  a )}2 = ( 3a  a ) 2  ( 3a  a ) 2 Now, –2x (a + b) – 2y (b – a) = – 2x (a – b) – 2y (a + b) Area of ABC = 3 (side)2 4  Area of ABC = 3 × AB2 4  Area of ABC = 3 × (2 2 a)2 sq. units 4  ax + bx + by – ay = ax – bx + ay + by  2bx = 2ay Ex.9 Sol.  bx = ay Find the value of x, if the distance between the points (x, – 1) and (3, 2) is 5. Let P(x, – 1) and Q(3, 2) be the given points, Then, PQ = 5 (Given)  ( x  3) 2  ( 1  2) 2 = 5  (x – 3)2 +9=  (x – 7) (x + 1) = 0  x = 7 or x = – 1 Ex.10 Show that the points (a, a), (–a, –a) and (– 3 a, 3 a) are the vertices of an equilateral triangle. Also find its area. Let A (a, a), B(–a, –a) and C(– 3 a, the given points. Then, we have Ex.11 Show that the points (1, – 1), (5, 2) and (9, 5) are collinear. Sol. 52  x2 – 6x + 18 = 25  x2 – 6x – 7 = 0 Sol. = 2 3 a2 sq. units 3 a) be Let A (1, –1), B (5, 2) and C (9, 5) be the given points. Then, we have AB = (5  1) 2  (2  1) 2 = 16  9 = 5 BC = (5  9) 2  ( 2  5) 2 = 16  9 = 5 and, AC = (1  9) 2  ( 1  5) 2 = 64  36 = 10 Clearly, AC = AB + BC Hence, A, B, C are collinear points. Ex.12 Show that four points (0, – 1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area. Sol. Let A (0, –1), B(6, 7), C(–2, 3) and D (8, 3) be the given points. Then, AD = (8  0) 2  (3  1) 2 = 64  16 = 4 5 BC = (6  2) 2  (7  3) 2 = 64  16 = 4 5 AC = ( 2  0) 2  (3  1) 2 = 4  16 = 2 5  SQ =  = a t 2 = (1  t 2 ) 2  4t 2 = a t2 a t 2 (1  t 2 ) 2 (1 + t2)  1 1 1 t2 + + = 2 2 SQ SP a ( t  1) a ( t  1)  1 1  t2 1 1 + = = , 2 SQ SP a a ( t  1) which is independent of t. and, BD = (8  6) 2  (3  7) 2 = 4  16 =2 5  AD = BC and AC = BD. Ex.14 If two vertices of an equilateral triangle be (0, 0), (3, 3 ), find the third vertex. So, ADBC is a parallelogram, Sol. O(0, 0) and A(3, 3 ) be the given points and let B(x, y) be the third vertex of equilateral OAB. Then, OA = OB = AB  OA2 = OB2 = AB2 Now, AB = (6  0) 2  (7  1)2 = and CD = 36  64 =10 (8  2) 2  (3  3) 2 = 10 Clearly, AB2 = AD2 + DB2 and CD2 = CB2 + BD2 Hence, ADBC is a rectangle. Now, Area of rectangle ADBC = AD × DB = (4 5 × 2 5 ) sq. units = 40 sq. units We have, OA2 = (3 – 0)2 + ( 3 – 0)2 = 12, OB2 = x2 + y2 Ex.13 If P and Q are two points whose coordinates are  a 2a  ,  respectively and S is 2 t  t 1 1 + is the point (a, 0). Show that SP SQ (at2, 2at) and  independent of t. Sol. =a ( t 2  1) 2  4 t 2 = a(t2 + 1) 2 and SQ =  SQ =  AB2 = x2 + y2 – 6x – 2 3 )2 3 y + 12  OA2 = OB2 = AB2  OA2 = OB2 and OB2 = AB2 (at 2  a ) 2  (2at  0) 2 We have, SP = and, AB2 = (x – 3)2 + (y – a   2a   0  2  a   t   t  a 2 (1  t 2 ) 2 t4  4a 2 t2 2  x2 + y2 = 12 and, x2 + y2 = x2 + y2 – 6x – 2  x2 + y2 = 12 and 6x + 2  x2 + y2 = 12 and 3x + 2  x2  6  3x   = 12 +   3  3 y + 12 3 y = 12 3y=6  (x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)2  6  3x   3x  3 y  6  y   3   x2 + y2 – 16x + 4y + 68 = x2 + y2 – 4x + 4y + 8  3x2 + (6 – 3x)2 = 36  12x = 60  12x2 – 36x = 0 x = 0, 3 So, the coordinates of the circumcentre P are (5, 2).  x=0  Also, Circum-radius = PA = PB = PC 3y=6 6 =2  y= 3 Putting x  0 in   3x  3y  6    3 and, x = 3  9 + 3 y = 6 y= 69 =– 3 Putting x  3 in   3x  3y  6    3 = (5  8) 2  (2  6) 2 = 9  16 = 5 Ex.16 If the opposite vertices of a square are (1, – 1) and (3, 4), find the coordinates of the remaining angular points. Sol. Let A(1, – 1) and C(3, 4) be the two opposite vertices of a square ABCD and let B(x, y) be the third vertex. Hence, the coordinates of the third vertex B are (0, 2 3 ) or (3, – 3 ). D Ex.15 Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). Also, find its circum radius. Sol.  x=5 C(1, –1) B(x, y) A(3, 4) Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A (8, 6), B(8, –2) and C(2, – 2) be the vertices of the given triangle and let P (x, y) be the circumcentre of this triangle. Then, Then, AB = BC  AB2 = BC2  (x – 1)2 + (y + 1)2 = (3 – x)2 + (4 – y)2  x2 – 2x + 1 + y2 + 2y + 1 = 9 – 6x + x2 + 16 – 8y + y2 PA = PB = PC  PA2 = PB2 = PC2  x2 + y2 – 2x + 2y + 2 = x2 + y2 – 6x – 8y + 25 A(8, 6)  4x + 10y = 23  x= P(x,y) B(8,–2) Now,  (x – 8)2 = + (y –   AB2 + BC2 = AC2  (x – 3)2 + (y – 4)2 + (x – 1)2 + (y + 1)2 = (3 –1)2 + (4 + 1)2 PB2 6)2 ….(1) In right-angled triangle ABC, we have C(2,–2) PA2 23  10 y 4 = (x – 8)2 + (y + 2)2  x2 + y2 – 16x – 12y + 100 = x2 + y2 – 16x + 4y + 68  x2 + y2 – 4x – 3y – 1 = 0 .... (2) Substituting the value of x from (1) and (2), we get 2  16y = 32  y = 2  23  10 y  2   + y – (23 – 10y) – 3y – 1 = 0 4   and, PB2 = PC2  4y2 – 12y + 5 = 0  (2y – 1) (2y – 5) = 0 1 5 or 2 2  y= Putting y = 1 5 and y = respectively in (1) we get 2 2 9 1 x= and x = respectively. 2 2 Hence, the required vertices of the square are 9 1  1 5  ,  and   ,  . 2 2    2 2 Ex.18 If P (2, – 1), Q(3, 4), R(–2, 3) and S(–3, –2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus. Sol. The given points are P(2, –1), Q(3, 4), R(–2, 3) and S(–3, –2). We have, PQ = QR = (3  2) 2  (4  1) 2 = 12  52 = 26 units (2  3) 2  (3  4) 2 = 25  1 = 26 units 26 units Ex.17 Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle. RS = (3  2) 2  (2  3) 2 = 1 25 = Sol. SP = (3  2) 2  (2  1) 2 = PR = (2  2)2  (3  1) 2 = 16  16 = 4 2 units Let A (–3, 0), B (1, –3) and C (4, 1) be the given points. Then, 2 2 26 units AB = {1  (3)}  (3  0) = 16  9 = 5 units. and, QS = 2 (4  1)  (1  3) BC = 2 = 9  16 = 5 units = & CA= (4  3) 2  (1  0) 2 = 49  1 =5 2 units y x'  (3  3) 2  (2  4) 2 36  36 = 6 2 units PQ = QR = RS = SP = 26 units and, PR  QS C(4,1) x A(–3,0) O B(1,3) y' Clearly, AB = BC. Therefore, ABC is isosceles. Also, AB2 + BC2 = 25 + 25 = (5)2 = CA2  ABC is right-angled at B. Thus, ABC is a right-angled isosceles triangle. Now, Area of ABC = 1 (Base × Height) 2 1 = (AB × BC) 2 1 2   Area of ABC =   5  5  sq. units = 25 sq. units. 2  This means that PQRS is a quadrilateral whose sides are equal but diagonals are not equal. Thus, PQRS is a rhombus but not a square. Now, Area of rhombus PQRS = 1 × (Product 2 of lengths of diagonals)  Area of rhombus PQRS = 1 × (PR × QS) 2  Area of rhombus PQRS 1  =   4 2  6 2  sq. units = 24 sq. units 2  Ex.19 Find the coordinates of the centre of the circle passing through the points (0, 0), (–2, 1) and (–3, 2). Also, find its radius. Sol. Let P (x, y) be the centre of the circle passing through the points O(0, 0), A(–2,1) and B(–3,2). Then, OP = AP = BP n B P m A If P is a point on AB produced such that AP : BP = m : n, then point P is said to divide AB externally in the ratio m : n. B(–3,2) m P B m P(x,y) n A A(–2,1) The coordinates of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are given by O(0, 0) Now, OP = AP  OP2 = AP2 mx 2  nx1 my 2  ny1   ,y x   mn mn    x2 + y2 = (x + 2)2 + (y – 1)2 y  x2 + y2 = x2 + y2 + 4x – 2y + 5  4x – 2y + 5 = 0 ....(1) B(x2, y2) n P(x, y) K m A(x1,y1) H and, OP = BP OP2 = BP2  x2 + y2 = (x + 3)2 + (y – 2)2  x2 + y2 = x2 + y2 + 6x – 4y + 13  6x – 4y + 13 = 0 L O ....(2) N M x The coordinates of P are On solving equations (1) and (2), we get  mx 2  nx1 my 2  ny1  ,   mn   mn 3 11 x= and y = 2 2 Note 1 :  3 11   2 2  Thus, the coordinates of the centre are  , x 2  y2 = Now, Radius = OP = =  SECTION FORMULAE 1 2 9 121  4 4 If P is the mid-point of AB, then it divides AB in the ratio 1 : 1, so its coordinates are  1 . x1  1. x 2 1. y1  1. y 2   x1  x 2 y1  y 2  , ,  =  11 11   2 2   Note 2 : 130 units. Fig. will help to remember the section formula. m  Let A and B be two points in the plane of the paper as shown in fig. and P be a point on the segment joining A and B such that AP : BP = m : n. Then, the point P divides segment AB internally in the ratio m : n. A(x1,y1) Note 3 : n P B(x2,y2) The ratio m : n can also be written as AQ : QB = 2 :  = 2 : 1 m : 1, or n m  : 1, where  = . n So, P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1. Thus, the coordinates of P and Q are So, the coordinates of point P dividing the line segment joining the points A(x1, y1) and B(x2, y2) are P  1  (3)  2  1 1  4  2  ( 2)   1  ,  = P  , 0 1 2 1 2    3   2  (3)  1  1 2  4  1  ( 2)   5  , , 2  =Q  2  1 2  1    3  m m  y 2  y1   x  x1  mx 2  nx1 my 2  ny1   n 2 n  , ,  = m  mn   m  mn 1 1   n  n   x 2  x1 y 2  y1  ,   1    1 = Q respectively Hence, the two points of trisection are (–1/3, 0) and (–5/3, 2). Type II : EXAMPLES  Type I : On finding the section point when the section ratio is given Ex.20 Find the coordinates of the point which divides the line segment joining the points (6, 3) and (– 4, 5) in the ratio 3 : 2 internally. Sol. Ex.22 In what ratio does the x-axis divide the line segment joining the points (2, –3) and (5, 6) ? Also, find the coordinates of the point of intersection. Sol. Let the required ratio be  : 1. Then, the coordinates of the point of division are, Let P (x, y) be the required point. Then,  5  2 6  3  R ,    1  1  3  ( 4)  2  6 3 5  2  3 x= and y = 3 2 3 2  x = 0 and y = On Finding the section ratio or an end point of the segment when the section point is given 21 5 So, the coordinates of P are (0, 21/5). Ex.21 Find the coordinates of points which trisect the line segment joining (1, –2) and (–3, 4). Sol. Let A(1, –2) and B(–3, 4) be the given points. Let the points of trisection be P and Q. Then, AP = PQ = QB =  (say).  A(1,–2) P   Q B(–3,4) PB = PQ + QB = 2 and AQ = AP + PQ = 2  AP : PB =  : 2 = 1 : 2 and But, it is a point on x-axis on which y-coordinates of every point is zero.  6  3 =0  1 = 1 2 Thus, the required ratio is 1 : 1 or 1 : 2. 2 Putting  = 1/2 in the coordinates of R, we find that its coordinates are (3, 0).   2  4 1  3  ,   = (1, 1) 2   2 Ex.23 If the point C (–1, 2) divides internally the line segment joining A (2, 5) and B in ratio 3 : 4, find the coordinates of B. Coordinates of the mid-point of BD are 11 0  2  ,   = (1, 1) 2   2 Let the coordinates of B be (, ). It is given that AC : BC = 3 : 4. So, the coordinates of C are Sol. Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, we shall see whether ABCD is a rectangle or not.  3  4  2 3  4  5   3  8 3  20  , ,   =  3 4   7 7   3 4 We have, AC = and, BD = But, the coordinates of C are (–1, 2)  Thus, the coordinates of B are (–5, –2). Ex.24 Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7). Sol. Suppose the line 3x + y – 9 = 0 divides the line segment joining A (1, 3) and B(2, 7) in the ratio k : 1 at point C. Then, the coordinates of C are  2k  1 7 k  3  ,    k 1 k 1  But, C lies on 3x + y – 9 = 0. Therefore, 7k  3  2k  1  3 –9=0  + k 1  k 1   6k + 3 + 7k + 3 – 9k – 9 = 0   k= 3 4 So, the required ratio is 3 : 4 internally. Type III : On determination of the type of a given quadrilateral Ex.25 Prove that the points (–2, –1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram. Is it a rectangle ? Sol. Let the given point be A, B, C and D respectively. Then, Coordinates of the mid-point of AC are (1  1) 2  (0  2) 2 = 2 Clearly, AC  BD. So, ABCD is not a rectangle. 3  8 3  20 = – 1 and =2 7 7  = – 5 and  = – 2 ( 4  (2)) 2  (3  ( 1)) 2 = 2 Ex.26 Prove that (4, – 1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square ? Sol. Let the given points be A, B, C and D respectively. Then, Coordinates of the mid-point of AC are  4  7  1  2   11 1  ,   = ,  2   2 2  2 Coordinates of the mid-point of BD are  6  5 0  1   11 1  ,   = ,  2   2 2  2 Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, AB = BC = (6  4) 2  (0  1) 2 = 5, (7  6) 2  ( 2  0) 2 = 5  AB = BC So, ABCD is a parallelogram whose adjacent sides are equal. Hence, ABCD is a rhombus. We have, AC = (7  4) 2  ( 2  1) 2 = 3 BD = (6  5) 2  (0  1) 2 = 2 and 2 Clearly, AC  BD. the mid-point of AC are same as the coordinates of the mid-point of BD i.e., So, ABCD is not a square.   2  4 1 b   a 1 0  2  , ,   =  2   2 2   2 Type IV : On finding the unknown vertex from given points Ex.27 The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.  1, Sol.  Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.  Coordiantes of the mid-point of AC = Coordinates of the mid-point of BD  1 2 0  2   3  x 1 y  , ,  =  2   2 2   2     1   3  x y 1 ,  , 1 =   2  2   2  3 x 1 y 1 = and =1 2 2 2  a + 1 = 2 and b – 1 = 2  a = 1 and b = 3 Ex.30 If the coordinates of the mid-points of the sides of a triangle are (1, 2) (0, –1) and (2, –1). Find the coordinates of its vertices. Sol. Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ABC. Let D (1, 2), E (0, –1), and F(2, –1) be the mid-points of sides BC, CA and AB respectively. Since D is the mid-point of BC.  x 2  x3 y  y3 = 1 and 2 =2 2 2  x2 + x3 = 2 and y2 + y3 = 4 Hence, the fourth vertex of the parallelogram is (–2, 1). Similarly, E and F are the mid-point of CA and AB respectively. We know that the diagonals of a parallelogram bisect each other. So, coordinates of the midpoint of diagonal AC are same as the coordinates of the mid-point of diagonal BD.   x1  x 3 y  y3 = 0 and 1 =–1 2 2  x1 + x3 = 0 and y1 + y3 = – 2 and .... (1) .... (2) x1  x 2 y  y2 = 2 and 1 =–1 2 2  6  9 1 4   8  p 2  3  , ,   =   2   2 2   2  x1 + x2 = 4 and y1 + y2 = –2  15 5   8  p 5  ,  =  ,  2  2 2  2 (x2 + x3) + (x1 + x3) + (x1 + x2) = 2 + 0 + 4 and,   8p 15  =  15 = 8 + p 2 2 p = 7 Ex.29 If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b. Sol. a 1 b 1 = 1 and =1 2 2  x = – 2 and y = 1 Ex.28 If the points A (6, 1), B (8, 2), C(9, 4) and D (p, 3) are vertices of a parallelogram, taken in order, find the value of p. Sol.  b 1  a 1  , 1  = 2   2   We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of .... (3) From (1), (2) and (3), we get (y2 + y3) + (y1 + y3) + (y1 + y2) = 4 –2 – 2  2(x1 + x2 + x3) = 6 and 2(y1 + y2 + y3) = 0  x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0 .... (4) A(7, –3) E(5, –2) F(6, 0) B(5, 3) C(3, –1) D(4, 1)  AD = (7  4) 2  (3  1) 2 = 9  16 = 5 units x1 + 2 = 3 and y1 + 4 = 0  x1 = 1 and y1 = – 4 So, the coordinates of A are (1, – 4) From (2) and (4), we get x2 + 0 = 3 and y2 – 2 = 0  x2 = 3 and y2 = 2 So, coordinates of B are (3, 2) From (3) and (4), we get (5  5) 2  (2  3) 2 = BE = From (1) and (4), we get (6  3) 2  (0  1) 2 = and, CF = So, coordinates of C are (–1, 2) Hence, the vertices of the triangle ABC are A(1, – 4), B(3, 2) and C(–1, 2). Ex.31 Find the lengths of the medians of a ABC whose vertices are A(7, –3), B(5,3) and C(3,–1). Sol. Let D, E, F be the mid-points of the sides BC, CA and AB respectively. Then, the coordinates of D, E and F are Sol. Let AD be the median through the vertex A of ABC. Then, D is the mid-point of BC. So, the   3 1  2  8  , coordinates of D are   i.e.,(–2, 3). 2   2  AD =  7  5 3 3 ,  = F (6, 0) 2   2 and, F  49  16 = 65 units Let G be the centroid of ABC. Then, G lies on median AD and divides it in the ratio 2 : 1. So, coordinates of G are  2  (2)  1  5 2  3  1  (1)  ,   2 1 2 1     4  5 6 1   1 ,  =  , 3  3  3 = 5  3 A(5, –1)  5  3 3 1 ,  = D(4, 1), 2   2  3  7 1 3  ,  = E (5, –2) 2   2 (5  2) 2  ( 1  3) 2 = D E 9  1 = 10 units. Ex.32 If A (5, –1), B(–3, –2) and C(–1, 8) are the vertices of triangle ABC, find the length of median through A and the coordinates of the centroid. x3 + 4 = 3 and y3 – 2 = 0  x3 = – 1 and y3 = 2 0  25 = 5 units G 2 1 B(–3, –2) D(–2, 3) C(–1, 8) APPLICATION OF SECTION FORMULA Theorem : The coordinates of the centroid of the triangle whose vertices are (x1, y1), (x2, y2) and  x1  x 2  x 3 y1  y 2  y3  ,  3 3   (x3, y3) are  From (1), (2) and (3), we get A(x1, y1)  x  x 2 y1  y 2  , F 1 2   2 G x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 6  x  x 3 y1  y 3  E 1 , 2   2 and, y1 + y2 + y2 + y3 + y1 + y3 = 2 – 6 + 8  x1 + x2 + x3 = 6 and y1 + y2 + y3 = 2...(4) The coordinates of the centroid of ABC are B(x2, y2)  x  x 3 y 2  y3  , D 2 2   2 C(x3, y3)  x1  x 2  x 3 y1  y 2  y 3   6 2  ,   =  ,  3 3   3 3  EXAMPLES  Ex.33 Find the coordinates of the centroid of a triangle whose vertices are (–1, 0), (5, –2) and (8, 2). Sol. We know that the coordinates of the centroid of a triangle whose angular points are (x1, y1), (x2, y2), (x3, y3) are  Sol. Let the coordinates of the third vertex be (x, y). Then, x 37 y54 = 2 and =–1 3 3  x – 4 = 6 and y – 1 = – 3 So, the coordiantes of the centroid of a triangle whose vertices are (–1, 0), (5, –2) and (8, 2) are  x = 10 and y = – 2  1 5  8 0  2  2  ,   or, (4, 0) 3 3   Thus, the coordinates of the third vertex are (10, –2). Ex.34 If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid. Let P (1, 1), Q(2, –3), R(3, 4) be the mid-points of sides AB, BC and CA respectively of triangle ABC. Let A (x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle ABC. Then, P is the mid-point of BC  [Using (4)] Ex.35 Two vertices of a triangle are (3, –5) and (–7, 4). If its centroid is (2, –1). Find the third vertex.  x1  x 2  x 3 y1  y 2  y 3  ,   3 3   Sol. 2 3 =  2,  Ex.36 Prove that the diagonals of a rectangle bisect each other and are equal. Sol. Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b. x1  x 2 y  y2 = 1, 1 =1 2 2  x1 + x2 = 2 and y1 + y2 = 2 ...(1) Q is the mid-point of BC  x 2  x3 y  y3 = 2, 2 =–3 2 2  x2 + x3 = 4 and y2 + y3 = – 6 ...(2) R is the mid-point of AC  Since, OACB is a rectangle. Therefore, x1  x 3 y y = 3 and 1 3 = 4 2 2  x1 + x3 = 6 and y1 + y3 = 8 Then, the coordinates of A and B are (a, 0) and (0, b) respectively. AC = ObAC = b ...(3) Thus, we have OA = a and AC = b So, the coordiantes of C are (a, b). The coordinates a 0 b0 ,   = 2   2 of the mid-point of OC are a b  ,  2 2 Also, the coordinates of the mid-points of AB a 0 0b a b are  ,  = ,  2  2 2  2 Clearly, coordinates of the mid-point of OC and AB are same. Hence, OC and AB bisect each other. Also, OC = a 2  b 2 and ( a  0) 2  ( 0  b ) 2 = AB = Remark : Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear iff Area of ABC = 0 i.e., x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0 EXAMPLES  Type I : On finding the area of a triangle when coordinates of its vertices are given. Ex.37 Find the area of a triangle whose vertices are A(3, 2), B (11, 8) and C(8, 12). Sol. Let A = (x1, y1) = (3, 2), B = (x2, y2) = (11, 8) and C = (x3, y3) = (8, 12) be the given points. Then, a 2  b2 Area of ABC =  OC = AB  AREA OF A TRIANGLE 1 |{x (y – y ) + x2 (y3 – y1) + x3 (y1 – y2)}| 2 1 2 3  Area of ABC = Theorem : The area of a triangle, the coordinates of whose vertices are (x1, y1), (x2, y2) and (x3, y3) is 1 | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) | 2 1 |{3(8 – 12) + 11 (12 – 2) + 8 (2 – 8)}| 2  Area of ABC = 1 |(–12 + 110 – 48)| = 25 sq. units 2 ALTER We have, Remark : The area of ABC can also be computed by using the following steps : Step I : Write the coordinates of the vertices A(x1, y1), B(x2, y2) and C(x3, y3) in three coloums as shown below and augment the coordinates of A(x1, y1) as fourth coloumn. Step II : Draw broken parallel lines pointing down wards from left to right and right to left. x1 x2 x3 x1 y1 y2 y3 y1 Step III : Compute the sum of the products of numbers at the ends of the lines pointing downwards from left to right and subtract from this sum the sum of the products of numbers at the ends of the lines pointing downward from right to left i.e., compute Area of ABC = |(3 × 8 + 11 × 12 + 8 × 2) – (11 × 2 + 8 × 8 + 3 × 12)| Area of ABC = 1 |(24 + 132 + 16) – (22 + 64 + 36)| 2 Area of ABC = 1 | 172 – 122 | = 25 sq. units 2 Ex.38 Prove that the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t. Sol.  (x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3) Step IV : Find the absolute of the number obtained in step III and take its half to obtain the area. Let A = (x1, y1) = (t, t – 2), B (x2, y2) = (t + 2, t + 2) and C = (x3, y3) = (t + 3, t) be the vertices of the given triangle. Then,   Area of ABC =  1 | x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)| 2  Area of ABC = 1 |{t(t + 2 – t) + (t + 2) (t – t + 2) 2 + (t + 3) (t – 2 – t – 2)}|  Area of ABC = 1 |{2t + 2t + 1 – 4t – 12}| = | – 4| 2 = 4 sq. untis Clearly, area of ABC is independent f t. ALTER  We have, t t+2 t+3 t t–2 t+2 t t–2 Area of DEF = Area of ABC 1 {t (t  2)  (t  2)t  (t  3)(t  2)}  {(t  2)(t  2) =  (t  3)(t  2)  t  t} 2 1 2 |(t + 2t + t2 + 2t + t2 + t – 6) 2  Area of ABC = – (t2 –4+ t2 + 5t + 6 + t2)| 1 | (3t2 + 5t – 6) – (3t2 + 5t + 2)| 2  Area of ABC =  Area of ABC = 1 | (–6 – 2)| 2  Area of ABC = 4 sq. units Hence, Area of ABC is independent of t. Ex.39 Find the area of the triangle formed by joining the mid-point of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of area of the triangle formed to the area of the given triangle. Sol. Let A (0, –1), B(2, 1) and C(0, 3) be the vertices of ABC. Let D, E, F be the midpoints of sides BC, CA and AB respectively. Then, the coordinates of D, E and F are (1, 2), (0, 1) and (1, 0) respectively. 1 |x (y – y ) + x2 (y3 – y1) + x3 (y1 – y2)| 2 1 2 3  Area of DEF = 1 |1 (1 – 0) + 0 (0 – 2) + 1 (2 – 1)| 2  Area of DEF = 1 |1 + 1| = 1 sq. units 2  Area of DEF : Area of ABC = 1 : 4 Ex.40 If D, E and F are the mid-points of sides BC, CA and AB respectively of a ABC, then using coordinate geometry prove that Area of DEF = Sol. 1 (Area of ABC) 4 Let A(x1, y1), B(x2, y2), C(x3, y3) be the vertices of ABC. Then, the coordinates of D, E and F  x 2  x 3 y 2  y3  ,   2   2 are  x1  x 3 y1  y3  ,   2   2  x1  x 2 y1  y 2  ,  respectively. 2   2 and  A(x1, y1) x  x y  y  F 1 2 , 1 2 2  2  x  x y  y  E 1 3 , 1 3 2  2  Now, Area of ABC = 1 | x (y – y ) + x2 (y3 – y1) + x3 (y1 – y2)| 2 1 2 3 Area of ABC = Area of ABC = 1 | 0(1 – 3) + 2 (3 – (–1)) + 0(0 – 1)| 2 1 | 0 + 8 + 0 | = 4 sq. units 2 B(x2, y2)  x 2  x 3 y 2  y 3  C(x3, y3) , D  2  2  1 = Area of ABC = 1 |x (y – y3) + x2 (y3 – y1) 2 1 2 + x3 (y1 – y2)| 2 = Area of DEF = A(4, 6) 1  x 2  x 3  y1  y3 y1  y 2   x1  x 3       2  2  2 2   2  1  y1  y 2 y 2  y 3  +  x1  x2   y2  y3 y1  y3         2   2  2 2   2 B(1, 5)  1  12 5  18   13 23   7  12 2  18   19  , ,  =  ,  =  , 5  and   1 3 1 3   4 4   1 3 1 3   4  respectively. + x3 (y3 – y2 + y1 – y3)|  1 1  2 = (Area of ABC) = 1 4 4  1 Hence, Area of DEF = (Area of ABC) 4 Ex.41 The vertices of ABC = are A (4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that AD 1 AE = = . Calculate the area of ADE AB 4 AC and compare it with the area of ABC. Sol. We have, AD 1 AE = = AB 4 AC  AB AC = =4 AD AE  AD  DB AE  EC = =4 AD AE We have, 4 6 DB EC AD AE 1 = = 3  = = AD AE DB EC 3  AD : DB = AE : EC = 1 : 3  D and E divide AB and AC respectively in the ratio 1 : 3. 13 19 4 4 23 5 4 6 4  Area of ADE = 1  23 13 19   13 19 23   4    5   6     6    4  5 2  4 4 4 4 4 4    Area of ABC =  1  92 65 114  78 437   20       2 4 4 4   4 16  1 271 1069  16 2 4  Area of ABC = = 1 15 15 = sq. untis. × 2 16 32 Also, we have DB EC  1+ =1+ =4 AD AE  C(7, 2) So, the co-ordinates of D and E are 1 |x (y – y3 + y2 – y1) + x2 (y3 – y2 + y2 – y1) 8 1 1 1 |x (y – y3) + x2 (y3 – y1) + x3 (y1 – y2)| 8 1 2 3 3 + (x1 + x2)(y2 – y1)|  2 = E D 1  2 = |(x2 + x3)(y3 – y2) + (x1 + x3)(y1 – y3) 8  2 = 1  4 1 7 4 6 5 2 6  Area of ABC = 1 |(4 × 5 + 1 × 2 + 7 × 6) 2 – (1 × 6 + 7 × 5 + 4 × 2)|  Area of ABC = 1 |(20 + 2 + 42) – (6 + 35 + 8)| 2  Area of ABC = 1 15 |64 – 49| = sq. units 2 2   Area of ADE 15 / 32 1 = = Area of ABC 15 / 2 16 TypeII : On finding the area of a quadrilateral when coordinates of its vertices are given Hence, Area of ADE : Area of ABC = 1 : 16. Ex.43 Find the area of the quadrilateral ABCD whose vertices are respectively A(1, 1), B(7, –3), C(12, 2) and D(7, 21). Ex.42 If A(4, –6), B(3, –2) and C(5, 2) are the vertices of ABC, then verify the fact that a median of a triangle ABC divides it into two triangle of equal areas. Sol. Let D be the mid-point of BC. Then, the coordinates of D are (4, 0). Sol. Area of quadrilateral ABCD = | Area of ABC | + | Area of ACD | We have,  Area of ABC = 1 |(1× –3 + 7 × 2 + 12 × 1) 2 – (7 × 1 + 12 × (–3) + 1× 2)| We have,  Area of ABC = 1 |(–3 + 14 + 12) – (7 – 36 + 2)| 2  Area of ABC = 1 |23 + 27| = 25 sq. units 2 Also, we have  Area of ABC = 1 |(4 × (– 2) + 3 × 2 + 5 × (– 6)) 2 – (3 × (– 6) + 5 × (–2) + 4 × 2)|  Area of ABC = 1 |( –8 + 6 – 30) – (–18 – 10 + 8)| 2  Area of ABC = 1 | –32 + 20 | = 6 sq. units 2 Also, We have  Area of ACD = 1 |(1 ×2 + 12 × 21 + 7 × 1) 2 – (12 × 1 + 7 × 2 + 1 × 21)|  Area of ACD = 1 |(2 + 252 + 7) – (12 + 14 + 21)| 2  Area of ACD = 1 |261 – 47| = 107 sq. units 2 Area of quadrilateral ABCD = 25 + 107 = 132 sq. 1 {(4  (2)  3  0  4  (6))}  Also of ABD = 2  {3  (6)  4  (2)  4  0} 1  Area of ABD = |(–8 + 0 + 26) – (–18 – 8 + 0)| 2  Area of ABD =  1 |(–32 + 26)| = 3 sq. units 2 Area of ABC 6 2 = = Area of ABD 3 1  Area of ABC = 2 (Area of ABD) units Type III : On collinearity of three points FORMULA: Three points (x1, y1), (x2, y2) and (x3, y3) are collinear iff x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0 or, (x1y2 + x2y3 + x3y1) – (x1y3 + x3y2 + x2y1) = 0 We have, Ex.44 Prove that the points (2, – 2), (–3, 8) and (–1, 4) are collinear. Let  be the area of the triangle formed by the given points. Sol. i.e., We have, |{2k2 + (–k + 1) (6 – 2k) + (–4 – k) (2 – 2k)} – {(–k + 1) (2 – 2k) + (–4 – k) (2k)  = + k (6 – 2k)}| = 0 1 |{2 × 8 + (–3) × 4 + (–1) × (–2)} 2       =  8k2 + 4k – 4 = 0  2k2 + k – 1 = 0  (2k – 1) (k + 1) = 0 1 |6 – 6| = 0 2  k = 1/2 or, k = – 1 Ex.45 Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear. Let  be the area of the triangle formed by the points (a, b + c), (b, c + a) and (c, a + b). We have, a b+c  = b c c+a a+b Hence, the given points are collinear for k = 1/2 or, k = – 1. Ex.47 For what value of x will the points (x, –1), (2, 1) and (4, 5) lie on a line ? Sol. Given points will be collinear if the area of the triangle formed by them is zero. a b+c x 2 4 x –1 1 5 –1  Area of the triangle = 0 1 |{a (c + a) + b (a + b) + c (b + c)} 2   |{x × 1 + 2 × 5 + 4 × (–1)} – {b (b + c) + c (c + a) + a (a + b)}| – {(2 × (–1) + 4 × 1 + x × 5}| = 0 1   = |(ac + a2 + ab + b2 + bc + c2) 2  (x + 10 – 4) – (–2 + 4 + 5x) = 0  – (b2 + bc + c2 + ca + a2 + ab)|  =0 Hence, the given points are collinear. Type IV : On Finding the desired result or unknown when three points are collinear Ex.46 For what value of k are the points (k, 2 – 2k) (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear ? Sol. + 2k2 + 6k – 8)  (6k2 – 2k – 2) – (–2k2 – 6k + 2) = 0 Hence, given points are collinear. Sol. + 6 –8k + 2k2 – (2 – 4k + 2k2 – 8k – 2k2 + 6k – 2k2)| – {(–3) × (–2) + (–1) × 8 + 2 × 4}| 1  = |(16 – 12 + 2) – (6 – 8 + 8)| 2 |(2k2 Given points will be collinear, if area of the triangle formed by them is zero.  (x + 6) – (5x + 2) = 0  – 4x + 4 = 0    x = 1 Hence, the given points lies on a line, if x = 1. Type V : Mixed problems based upon the concept of area of a triangle Ex.48 If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the coordniates of any point P, if Sol. PA = PB and Area of PAB = 10. Show that the ratio of the areas of triangle PBC Let the coordinates of P be (x, y). Then, and ABC is PA = PB Sol.  PA2 = PB2 We have,  (x – 3)2 + (y– 4)2 = (x – 5)2 + (y + 2)2  x – 3y – 1 = 0 ....(1) Now, Area of PAB = 10  x 3 5 x y 4 –2 y 1 |(4x + 3 × (–2) + 5y) – (3y + 20 – 2x)| = 10 2  |(4x + 5y – 6) – (–2x + 3y + 20)| = 20 |6x + 2y – 26| = ± 20  6x + 2y – 26 = ± 20  6x + 2y – 46 = 0 or, 6x + 2y – 6 = 0 xy2 . 7 x –3 4 x y 5 –2 y   Area of PBC = 1 |(5x+6+4y)–(–3y+20–2x)| 2   Area of PBC = 1 |5x + 6 + 4y + 3y – 20 + 2x| 2   Area of PBC = 1 |7x + 7y – 14| 2   Area of PBC = 7 |x + y– 2| 2   Area of PBC = 7 |6 + 3 – 2| 2  3x + y – 23 = 0 or, 3x + y – 3 = 0 Re placing x by 6 and y  3 in Area of PBC    Solving x – 3y – 1 = 0 and 3x + y – 23 = 0 we get x = 7, y = 2. Solving x – 3y – 1 = 0 and 3x + y – 3 = 0, we get x = 1, y = 0.  Area of ABC =  7 |xy2| Area of PBC  = 2 49 Area of ABC 2 Thus, the coordinates of P are (7, 2) or (1, 0). Ex.49 The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). 49 2  = xy2 |xy2| = 7 7 IMPORTANT POINTS TO BE REMEMBERED Ordinate 1. The point of intersection of two perpendicular lines is known as origin O(0, 0) . 2. The Horizontal line XOX' is called x-axis and the vertical line YOY' is call y-axis. X' 4. These two mutually perpendicular lines divide the cartesian plane into four parts which are called quadrants I, II, III and IV. 5. Quadrant 1st 2nd 3rd 4th Abscissa(x) + – – + Ordinate(y) + + – – Co-ordinate(x,y) (+, +) (–, +) (–, –) (+, –) 6. In a point A(x, y), x-cordinate is known as Abscissa and y-coordinate is known as ordinate. 7. Equation of x-axis is y = 0 Ordinate 3. The plane on which these lines are graphed is known as cartesian Plane and lines are called the co-ordinate axes. 12. Abscissa is +ve to the right of the origin and negative to the left of the origin. 13. Ordinate (y) is +ve above x-axis and –ve below x-axis. 14. Sign convention- While marking a point on the squared paper, the sign convention must be kept in mind. Y 4 3 2 I Quadrant II Quadrant 1 (–, +) (+, +) X' –4 –3 –2 –1 III Quadrant (–, –) Y' O 1 2 3 4 X –1 –2 IV Quadrant –3 (+, –) –4 X 15. Distance between two points (x1, B(x2, y2) is AB = 2 ( x 2  x1 )  ( y 2  y1 ) y1) and 2 16. Distance of a point P(x, y) from the origin O(0,0) isOP = ( x  0) 2  ( y  0) 2 = x 2  y2 17. Section formula : the co-ordinates of the point P(x, y) which divide the join of the points A(x, y) and B(x2, y2) internally in the ratio m1 : m2 are  m1x 2  m 2 x1 m1y 2  m 2 y1    ,  m m m1  m 2  1 2  9. Graph of the line x = a is always parallel to y-axis. 11. Any point on x-axis is (x, 0) and any point on y-axis is (0, y) Abscissa O (Origin) Y' 8. Equation of y-axis is x = 0 10. Graph of the line y = b is always parallel to x-axis. Abscissa Y 18. The mid-point of a line segment divides the line segment in the ratio 1:1. Therefore, the co-ordinates of the point P of the join of the points A(x1, y1) and B(x2,y2) are  x1  x 2 y1  y 2  ,   2   2 19. Area of ABC whose vertices B(x2, y2) and C(x3, y3) are area of ABC = A(x1, y1), 1 [x (y – y3) + x2(y3 – y1) 2 1 2 + x3(y1 –y2)] OR x1 x2 x3 x1 y1 y2 y3 y1 OR area of ABC = 1 [(x1y2 + x2y3 + x3y1) 2 – (x1y3 + x3y2 + x2y1)] 20. The points A, B, C are collinear if area of ABC = 0.  x1  x 2  x 3 y1  y 2  y 3  ,  3 3   21. Centroid :  22. AREA OF QUADRILATERAL If A (x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) be the vertices of quadrilateral ABCD then area of the quadrilateral ABCD = 1 [x y –x y + x2y3 –x3y2 + x3y4 – x4y3 + 2 1 2 2 1 x4y1 – x1y4]