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POINT, , POINT, 1. SYSTEM OF CO-ORDINATES, , 1.1, , Let ‘O’ be any fixed point in a plane and XOX’ and YOY’ be two perpendicular lines drawn through any, point O in the plane of the paper. Then, , , , The line XOX’ is called x-axis., , , , The line YOY’ is called y-axis., , , , Co-ordinate axes: x-axis and y-axis together are called axes of co-ordinates or axes of reference., , , , Origin: The fixed point ‘O’ is called the origin of co-ordinates or the origin., , , , Oblique axis: If both the axes are not perpendicular then they are called as oblique axes., , , , Cartesian Co-ordinates: The ordered pair of perpendicular distance from both axes of a point P lying in the, plane is called Cartesian Co-ordinates of P. If the cartesian co-ordinates of a point P are (x, y) then x is, called abscissa or x co-ordinate of P and y is called the ordinate or y co-ordinate of point P., Y, , Note :, (i), co-ordinates of the origin is (0, 0), (ii), , x, , y co-ordinate on x-axis is zero., , P(x,y), y, , X', , (iii), , x co-ordinate on y-axis is zero., , 1.2, , Polar Co-ordinates: Let OX be any fixed line which is usually, called the intial line and O be a fixed point on it. If distance of any, , X, , O, , Y', , point P from the origin O is ‘r’ and XOP   then  r,  are, called the polar co-ordinate of a point P. If (x, y) are the cartesian, , Y, , Pr, , co-ordinates of a point P, then x  r cos , y  r sin  and, , r, , y, r  x 2  y 2 ,   tan1  , x, , , O, , X, , 2. DISTANCE FORMULA, The distance between two points P  x1,y1  and Q  x 2 ,y 2  is given by PQ , Note :, (i) In particular the Distance of a point P  x,y  from the origin , ,  x1  x 2 , , 2, ,   y1  y 2 , , 2, , x2  y2, , (ii) Distance between two polar co-ordinates A(r1 , 1 ) and B(r2 , 2 ) is given by AB  r12  r22  2r1 r2 cos  1  2 , Ex.1, , The distance between P  3, 2  and Q  7, 5  is, [1], , Sol., , Ex.2, , Sol., , [2], , 115, , PQ , , 3  7 , , 2, , 109, , [3], , 91, , [4] 11, ,   2  5   100  9  109, 2, , , ,  , The distance between P  2,   and Q  3,  is, 6, ,  6, [1] 3, [2] 1/2, , Ans. [2], , [3], , 7, , 1,   ,  , PQ  22  32  2.2.3 cos      13  12  cos     13  12   7, 2,  6 6,  3, , [4], , 5, Ans. [3], , 2

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POINT, , 3. APPLICATION OF DISTANCE FORMULAE, (i), , For given three points A, B, C to decide whether they are collinear or vertices of a particular triangle. After finding, AB, BC and CA we shall find that the point are:, , Ex.3, , , , Collinear: If the sum of any two distances is equal to the third., , , , Vertices of an equilateral triangle if AB = BC = CA, , , , Vertices of an isosceles triangle if AB = BC or BC = CA or CA = AB, , , , Vertices of a right angled triangle AB2  BC2  CA 2 etc.,, , If A  2, 2  ; B  2,1 and C  5,2  are three points then A, B, C are, [1] collinear, , [2] vertices of an equilateral triangle, , [3] vertices of right angled triangle, , [4] none of these, , Sol., , AB, , , ,  2  2 , , 2, ,  1  2 , , BC, , , ,  5  2, , 2, ,   2  1, , CA , , 2  5, , 2, ,   2  2 , , 2, , 2, ,  5,  5 2, , 2, ,  5, , Since the sum of any two distances is not equal to the third so A, B, C are not collinear. They are vertices of a, triangle. Also AB2  CA 2  BC2, ,  A,B,C are vertices of a right angled triangle., (ii), , Ans. [3], , For given four points:, , , AB  BC  CD  DA; AC  BD  ABCD square, , , , AB  BC  CD  DA; AC  BD  ABCD rhombus, , , , AB  CD, BC  DA, AC  BD  ABCD is a rectangle, , , , AB  CD, BC  DA, AC  BD  ABCD is a parallelogram, QUADRILATERAL, , DIAGONALS, , ANGLE BETWEEN DIAGONALS, , (i), , Parallelogram, , Not equal, , , , , 2, , (ii), , Rectangle, , Equal, , , , , 2, , (iii), , Rhombus, , Not equal, , , , , 2, , (iv), , Square, , Equal, , , , , 2, , Note :, (i), , Diagonal of square, rhombus, rectangle and parallelogram always bisect each other., , (ii), , Diagonal of rhombus and square bisect each other at right angle., , (iii), , Four given points are collinear, if area of quadrilateral is zero., 3

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POINT, , Ex.4, , Sol., , Points A 1,1 ,B  2,7  and C  3, 3  are, [1] collinear, , [2] vertices of equilateral triangle, , [3] vertices of isoscele triangle, , [4] none of these, , AB , , 1  2, , BC , ,  2  3 , , CA , ,  3  1, , 2, ,  1  7   9  36  3 5, , 2, , 2, , 2, ,   7  3   25  100  5 5, 2, ,   3  1  4  16  2 5, 2, , Clearly BC  AB  AC . Hence A,B,C are collinear.., , Ans. [1], , 4. SECTION FORMULA, Co-ordinates of a point which divides the line segment joining two points P  x1,y1  and Q  x 2 ,y 2  in the ratio, m1 : m2 are, (1), ,  m1x 2  m2 x1 m1y 2  m2 y1 , , For internal division   m  m , m  m, , 1, 2, 1, 2, , , (2), ,  m1x 2  m2 x1 m1y 2  m2 y1 , , For external division   m  m , m  m, , 1, 2, 1, 2, , , (3), ,  x 1  x 2 y1  y 2 , ,, Co-ordinates of mid point of PQ , Put m1  m2, 2 ,  2, , (4), , Co-ordinates of any point on the line segment joining two points P(x1, y1) and Q  x 2 ,y 2  are,  x 1   x 2 y1   y 2 ,    1 , 1    ,   1, , , , 4.1 DIVISION BY AXES: PQ is divided by, (i), 4.2, Ex.5, ,  y1, x-axis in the ratio  y, 2, , x1, (ii) y-axis in the ratio   x, 2, , ax1  by1  c, DIVISION BY A LINE: A line ax  by  c  0 divides PQ in the ratio  ax  by  c, 2, 2, , The co-ordinates of point of internal and external division of the line segment joining two points  3, 1 and  3,4 , in the ratio 2 : 3 are respectively, [1] (2, 3) (11, 3), , Sol., , [2] (3, 1) (3, –11), , [3] (1, 3) (–11, 3), , [4] (1, –3) (11, –3), , Internal division, , x, , 2 3  3 3, , 23, External division, x, , 2 3  3 3, 23, , 3, , y, , 3, , y, , 2  4   3  1, 23, 2  4   3  1, 23, , 1, , Hence point  3,1, ,  11 Hence point  3, 11, , Ans.[2], , 4

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POINT, , Ex.6, , The ratio in which the line 3x  4y  7 divides the line segment joining the points 1,2  and  2,1 is, [1], , 4, 9, , [2], , 9, 4, , [3], , 1, 3, , Sol., , 3 1  4  2   7, ax1  by1  c, 4, 4, Required ratio =  ax  by  c   3 2  4 1  7   9  9, , , , , 2, 2, , Ex.7, , The points of trisection of line joining the points A  2,1 and B  5,3  are, ,  5  7, [1]  3,  ,  4, ,  3  3, , Sol., ,  2,1, ,  1  2 , x, A P1 ', P2 ', ,  3  3 , [2]  3,  4, ,  5  7 , , 5 , 7, , [3]  3,  4,  , 3 , 3, , , [4], , 3, 4, , Ans. [1], , 5  3 , , [4]  3,    4, , 3  7 , , , x  5,3 , ,  2  1 , ,  1 5  2  2 1 3  2  1   5 , P1  x,y   , ,,  3,, 1 2, 1  2   3 , , ,  2  5  1 2 2  3  1 1   7 , P2  x,y   , ,,  4,, 2 1, 2  1   3 , , , Ex.8, , The ratio in which the lines joining the  3, 4  and  5,6  divided by x-axis is, [1] 3 : 2, , Sol., , Ans.[1], , [2] 2 : 3, , [3] 4 : 3, , y1,  4 , Required ratio =  y   ,   2:3,  6 , 2, , [3] 3 : 4, Ans.[2], , 5. CO-ORDINATES OF SOME PARTICULAR POINTS:, Let A  x1,y1  , B  x 2 ,y 2  and C  x 3 ,y3  are vertices of any triangle ABC, then, 5.1, , CENTROID:, The centroid is the point of intersection of the medians (line joining the mid point of sides and opposite vertices)., Centroid divides the median in the ratio of 2:1.,  x1  x 2  x 3 y1  y 2  y 3 , ,, Co-ordinates of centroid G , , 3, 3, , , A, , F, , 2, , E, G, , B, , 1, D, , C, 5

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POINT, , 5.2, , INCENTRE:, The incentre of the point of intersection of internal bisector of the angle. Also it is a centre of a circle touching all, the sides of a triangle., , A, ,  ax1  bx 2  cx 3 ay1  by 2  cy 3 , ,, Co-ordinates of incentre , , abc, abc, , , Where a, b, c are the sides of triangle ABC., , F, , NOTE:, , 5.3, , B, , (i), , Angle bisector divides the opposite sides in the ratio of remaining sides., BD AB c, , , Eg:, DC AC b, , (ii), , Incentre divides the angles bisectors in the ratio  b  c  : a,  c  a  : b,  a  b  : c, , I, , E, , C, , D, , CIRCUMCENTRE:, It is the point of intersection of perpendicular bisectors of the sides of a triangle. If O is the circumcentre of any, triangle ABC, then OA 2  OB2  OC2, A (x1, y)1, , D, , O, , F, , (x2,y)2, B, , E, , C, , (x3,y)3, , Note :, (i) If a triangle is right angle, then its circumcentre is mid point of hypotenuse., , 5.4, ,  x1 sin 2A  x2 sin 2B  x3 sin 2C y1 sin2A  y2 sin 2B  y3 sin 2C , ,, (ii) Cicumcentre is , , sin 2A  sin 2B  sin 2C, sin 2A  sin 2B  sin2C, , , ORTHOCENTRE:, , It is the point of intersection of perpendicular drawn from vertices on opposite sides of a triangle and can be, obtained by solving the equation of any two altitudes., A (x1 , y1), , D, (x2 , y2), B, , Note :, , O, , E, , (x3 , y3), C, , (i) If a triangle is right angle triangle, then orthocentre is the point where right angle is formed.,  x 1 tan A  x 2 tan B  x 3 tan C y1 tan A  y 2 tan B  y 3 tan C , ,, , (ii) Orthocentre is , tan A  tan B  tan C, tan A  tan B  tan C, , , REMARKS :, (i), (ii), (iii), , If the triangle is equilateral, then centroid, incentre, orthocentre, circumcentre, coincides., Orthocentre, centroid and circumcentre are always collinear and centroid divides the line joining., orthocentre and circumcentre in the ratio 2 : 1, In an isosceles triangle centroid, orthocentre, incentre, circumcentre lies on the same line., 6

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POINT, , 5.5, , EX-CENTRES :, The centre of the circle which touches side BC and the extended portions of sides AB and AC is called the excentre of ABC with respect to the vertex A. It is denoted by I1 and its coordinates are,  ax1  bx 2  cx 3 ay1  by 2  cy 3 , ,, I1  , , a  b  c, a  b  c, , , , A, , Similarly ex-centres of ABC with respect to vertices B and C are denoted by I2 and, I3 respectively , and, , I3, , I2, , B, , C, , I1, ,  ax  bx 2  cx 3 ay1  by 2  cy 3 , I2   1, ,, , a b  c, abc, , ,  ax  bx 2  cx 3 ay1  by 2  cy 3 , I3   1, ,, , ab c, abc, , , , Ex.9, , Centroid of the triangle whose vertices are  0,0  ,  2,5  and  7,4  is, [1] (4, 3), , [2] (3, 4), , [3] (3, 3), , Sol., , 027 05 4, ,, ,    3,3 , 3, 3, , , , Ex.10, , Incentre of triangle whose vertices are A  36,7  , B  20,7  , C  0, 8  is, , Ans.[3], , [1] (1, 1), Sol., , [4] (3, 5), , [2] (0, –1), , [3] (–1, 0), , [4] (1, 0), , Using distance formula, a  BC  202   7  8   25 , , b  CA  362   7  8   39, , 2, , c  AB , ,  36  20 , , 2, , 2, ,  25  36   39  20   56  0  25  7   39  7   56  8  , I  , ,, , 25  39  56, 25  39  56, , , ,   7  7   56 ,, 2, , I   1,0 , Ex.11, , Ans.[3], , If 1,4  is the centroid of a triangle and its two vertices are  4, 3  and  9,7  then third vertices is, [1] (7, 8), , Sol., , Ex.12, , [3] (8, 7), , [4] (6, 8), , Let the third vertices of triangle be  x,y  then, x49, 1, x8, 3, y37, 4,  y  8 So third vertex is (8, 8)., 3, , Ans.[2], , If  0,1 , 1,1 and 1,0  are middle points of the sides of a triangle, then find its incentre is, , , , [1] 2  2 , 2  2, Sol., , [2] (8, 8), , , , , , [2] 2  2,  2  2, , , , , , [3] 2  2 , 2  2, , , , , , [4] 2  2 ,  2  2, , , , Let A  x1,y1  , B  x 2 ,y 2  and C  x 3 ,y3  are vertices of a triangle, then, , x1  x 2  0, x 2  x3  2, x 3  x1  2, y1  y 2  2, y 2  y3  2, y 3  y1  0, Solving these equations, we get A  0,0  , B  0,2  and C  2,0  Now a  BC  2 2, b  CA  2, c  AB  2, , , , Thus incentre of ABC is 2  2, 2  2, , , , Ans.[1], 7

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POINT, , Ex. 13 Two vertices of a triagle are ( 5, –1) and ( –2, 3). If origin is the orthocentre, then the third vertex of the triangle is, (1) ( 4, –7), Sol., , (2) (–4, 7), , (3) (–4, –7), , (4) ( 4 , 7), , Let C (  , ) be the third vertex, A(5, –1), ,    3   1 , AO  BC  ,    1,    2  5 , ,  5 –  = – 13, , ....(1), ,    1  3 , BO  AC  ,     1,    5  2 , ,  2 –3 = 13, , O(0,0), , ..... (2), , B(–2, 3), , C(,), , Solving (1) and (2) , ( , ) = ( –4 , –7), , Ex.14, , If O  0,0  ; A  3,0  and B  0,4  are vertices of a triangle, then its circumcentre is, 3 , [2]  , 2 , 2 , , [1] (1, 1), Sol., , Ans.[3], ,  4, [3] 1, ,  3, ,  3, [4]  2, , 2, , , Let it be P  x,y  . Then PO2  PA 2  PB2,  x2  y 2   x  3   y2  x2   y  4 , 2, ,  0  6x  9  8y  16, , x, , 2, , 3, ,y2, 2, , 3 ,  circumcentre   2 , 2 , , , , Ans.[2], , 6. AREA OF TRIANGLE AND QUADRILATERAL, 6.1, , AREA OF TRIANGLE (Cartesian Coordinates), Let A  x1,y1  , B  x 2 ,y 2  and C  x 3 ,y3  are vertices of a triangle, then, , x1, 1, Area of triangle ABC  x 2, 2, x3, , y1 1, y2 1 , y3 1, , 1,  x1  y 2  y 3   x 2  y3  y1   x 3  y1  y 2  , 2, , Condition of collinearity:, Three points A  x1,y1  , B  x 2 ,y 2  , C  x 3 ,y 3  are collinear if the area of ABC  0 i.e.,, , if, , x1, x2, , y1 1, y2 1  0, , x3, , y3 1, , Particular cases:, , 1, x1y 2  x 2 y 1, 2, , (i), , When one vertex is origin i.e., if the vertices are  0,0  ;  x1, y1  and  x 2 , y 2  then its area  , , (ii), , When abscissae or ordinates of all vertices are equal then its area is zero., , (iii), , When two vertices be on x-axis say  a,0  , b,0  and third vertex be  h,k  , then its area , , 1, ab k, 2, 8

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POINT, , 1, cdh, 2, , (iv), , When two vertices be on y-axis say  0,c  ,  0,d and third vertex be  h,k  , then its area , , (v), , Area of the triangle formed by coordinate axes and the line ax  by  c  0 is, , (vi), , When ABC is right angled triangle and B  900 , then  , , (vii), , When ABC is equilateral triangle, then  , , (viii), , When D, E, F are the mid points of the sides AB, BC, CA of the triangle ABC, then its area   4  DEF , , c2, 2ab, , 1,  AB  BC , 2, , 3, 1, 2, 2,  side   height , 4, 3, , Note: Area of a triangle is always taken to be non-negative. So always use mod sign while using area formula., Ex.15, , If the vertices of a triangle are 1,2  4, 6  and  3,5  then its area is, [1], , 25, sq. unit, 2, , [2] 12 sq. unit, , [3] 5 sq. unit, , 1, 1, 25, 1 6  5   4  5  2   3  2  6     11  12  24  , square unit, 2, 2, 2, , Sol., , , , 6.2, , AREA OF QUADRILATERAL, , [4] 25 sq. unit, , Ans.[1], , If  x1,y1  ,  x 2 ,y2  ,  x 3 ,y3  and  x 4 ,y 4  are vertices of a quadrilateral then its area, , , , 1,  x1y 2  x 2 y1    x 2 y3  x3 y 2    x 3 y 4  x 4 y 3    x 4 y1  x1y 4 , 2, , Note:, (i), , If the area of quadrilateral joining four points is zero then those four points are collinear., , (ii), , If two opposite vertex of rectangle are  x1, y1  and  x 2 , y 2  then its area may be   y 2  y1  x 2  x1 , , (iii), , If two opposite vertex of a square are A  x1, y1  and C  x 2 ,y 2  then its area is, , , Ex.16, , If 1,1 3,4  5, 2  and  4, 7  are vertices of a quadrilateral then its area is, [1], , Sol., , Ex.17, , Sol., , 1, 1, 2, 2, AC2   x 2  x1    y 2  y1  , , , 2, 2, , , , 41, sq. units, 2, , [2] 41 sq. units, , [3] 20 sq. units, , 1, 1(4)  3(1)  3  2   5  4   5  7   4  2   4 1  1 7  , 2, 1, 41,   4  3  6  20  35  8  4  7 , sq. units, 2, 2, , [4] 22 sq. units, , Ans.[1], , If the coordinates of two opposite vertex of a square are  a,b  and  b,a  then area of square is, [1] (a + b)2, , [2] 2(a + b)2, , We know that area of square , , 1 2, d, 2, , , , 1, 2, 2, 2,  a  b    b  a     a  b , 2, , [3] (a – b)2, , [4] 2(a – b)2, , Ans.[3], 9

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POINT, , 6.3, , AREA OF A TRIANGLE (Polar Coordinates), If  r1, 1  ,  r2 , 2  and  r3 ,3  are vertices of a triangle then its area  ., , , , Ex.18, , 1, r1r2 sin  2  1   r2r3 sin  3  2   r3r1  1  3  , 2, ,  , 2 , , The area of a triangle with vertices  a,   ;  2a,    ;  3a,  , 3 , 3 , , [1], , 5 2 2, a, 4, , [2], , 2 5 2, a, 4, , [3], , 5 3 2, a, 4, , [4], , , , 1 2,  2   5 3 2, 2a sin  6a2 sin  3a2 sin , a,  , , 2, 3, 3, 4,  3 , , Sol., , , , 6.4, , AREA OF A TRIANGLE WHEN EQUATIONS OF ITS SIDES ARE GIVEN:, , 2 3 2, a, 4, , Ans.[3], , If ar x  br y  cr  0  r  1,2,3  are sides of a triangle then its area is given by, a1, 1, , a2, 2C1C2C3, a3, , b1, b2, b3, , c1, c2, c3, , 2, , when C1, C2, C3 are cofactors of c1,c 2 ,c 3 in the determinant., Ex.19, , If x – y = 1, x + 2y = 0 and 2x + y = 3 are sides of a triangle, then its area is, [1], , 2, 3, , [2], , 3, 2, , 1 1 1, , Sol., , 1, Area = 2( 3)( 3)3 1, 2, , 2, , 0, , 1, , 3, , [3] 2, , [4], , 1, 2, , 2, , , , 1, 2,  36  ., 54, 3, , Ans.[1], , 7. LOCUS OF A POINT, The locus of a moving point is the path traced out by that point under one or more given conditions., How to find the locus of a point : Let (x1, y1) be the co-ordinate of the moving points say P. Now apply the, geometrical conditions on x1, y1. This gives a relation between x1, and y1. Now replace x1 by x and y1 by y in the, eleminant and resulting equation would be the equation of the locus., Note :, (i), , Locus of of a point P which is equidistant from the two point A and B is straight line and is a perpendicular, bisector of line AB., , (ii), , In above case if, , PA  KPB where K  1 then the locus of P is a circle., (iii), , Locus of P if A and B are fixed., (a), , Circle if APB  cons tan t, , (b), , Circle with diameter AB if APB , , (c), , Ellipse if PA  PB  cons tan t, , (d), , Hyperbola if PA  PB  cons tan t, , , 2, , 10

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POINT, , Ex.20, , The locus of a point such that the sum of its distances from the points  0,2 and  0, 2  is 6 is, [1] 9x2 + 5y2 = 45, , Sol., , [2] 5x2 + 9y2 = 45, , [3] 4x2 + 7y2 = 35, , [4] 9x2 + 5y2 = 50, , Let P h,k  be any point on the locus and let A  0,2  and B  0, 2  be the given points., By the given condition PA  PB  6, , , h  0 , , 2, ,  k  2  , 2, , h  0 , , 2, ,  k  2   6, 2, ,  h2   k  2   6  h2   k  2 , 2, , 2, ,  h2   k  2   36  12 h2  k  2   h2   k  2 , 2, ,  8k  36  12 h2   k  2 , , 2, , 2, , 2, , , ,   2k  9   9 h2   k  2 , 2, , 2, , , ,  4k 2  36k  81  9h2  9k 2  36k  36  9h2  5k 2  45, Hence, locus of  h,k  is 9x 2  5y 2  45, , Ans.[1], , 11

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POINT, , SOLVED EXAMPLES, Ex.1, , The distance between the point P  acos , a sin   and Q  acos , a sin   is, (1) 4a sin, , Sol., ,  , 2, , (2) 2a sin, , , 2, , (3) 2a sin, ,  , 2, , (4) 2acos, , d2   a cos   a cos     a sin   a sin    a2  cos   cos    a2  sin   sin  , 2, , 2, , 2, , 2, ,  , 2, , 2, , 2, , ,   , ,   , , 2 ,  a2 2 sin, sin, sin,   a 2cos, , 2, 2, 2, 2 , , , , ,  4a2 sin2, Ex.2, ,  2  ,   , ,  cos2, sin,   4a2 sin2, 2 , 2, 2 , 2, , , 2, , Ans.[3], , The number of points on x-axis which are at a distance c  c  3  from the point (2, 3) is, (1) 2, , Sol., ,  d  2a sin, , (2) 1, , (3) infinite, , (4) no point, , Let a point on x-axis is  x1,0  then its distance from the point  2,3 , , ,  x1  2 , , 2, , or  x1  2   c 2  9, 2, , 9 c, ,  x1  2  c 2  9, , But c  3  c 2  9  0, ,  x1 will be imaginary, Ex.3, , If A 1,4  ; B  3,0  and C  2,1 are vertices of a triangle then the length of the median through C is, (1) 1, , Sol., , Ans.[4], , (2) 2, , (3), , 2, , (4), , 3, , The mid point D of side AB   2,2 , ,  the length of the median CD, , , Ex.4, ,  2  2, , 2, ,   2  1  1, 2, , Ans.[1], , The vertices of a triangle are A  0, 6  , B  6,0  and C 1,1 respectively, then coordinates of the ex-centre, opposite to vertex A is, ,  3 3 , (1)  , ,  2 2 , Sol., , 3, , (2)  4, , 2, , , a  BC , ,  6  1, , 2, , b  CA , , 1  0 , ,  1  6   50  5 2, , c  AB , , 0  6, , 2, , 2, ,  3 3 , (3)  , ,  2 2, , (4)  4, 6 , ,   0  1  50  5 2, 2, , 2, ,   6  0   72  6 2, 2, , coordinates of ex-centre opposite to vertex A are, , x, , ax1  bx 2  cx 3 5 2.0  5 2  6   6 2 1 24 2, , ,  4, a  b  c, 6 2, 5 2  5 2  6 2, , y, , ay1  by 2  cy 3 5 2  6   5 2.0  6 2 1 36 2, , ,  6, a  b  c, 6 2, 5 2  5 2  6 2, , Hence coordinates of ex-centre are  4, 6 , , Ans.[4], 12

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POINT, , Ex.5, , The point A divides the join of the points  5,1 and  3,5  in the ratio k:1 and coordinates of points B and C are, , 1,5 , , and  7, 2  respectively. If the area of ABC be 2 units, then k equals, , (1) 7, 9, Sol., , (3) 7,, , (2) 6, 7, , 31, 9, , (4) 9,, , 31, 9, ,  3k  5 5k  1 , A, ,, ,  k 1 k 1 , 1  3k  5, 5k  1 ,  5k  1, ,  5  2   1 2 ,   7  k  1  5    2, , 2  k  1, k, 1, , , , , , Area of ABC  2 units, , , ,  14k  66  4 k  1, ,  k  7 or, , 31, 9, , Ans.[3], , Ex.6, , 2, 2, The condition that the three points  a,0  ,  at1 ,2at1  and  at 2 ,2at 2  are collinear if, , Sol., , (2) t1t 2  2, (3) t1t 2  1, (1) t1  t 2  0, Here the points are collinear if the area of the triangle is zero., , (4) None of these, , 1 2, a t1  1 2at 2  2at1 at 22  a   0, 2, , , , Hence, , , , , , 2, 2, or t 2  t1  1  t1  t 2  1  0, , , ,   t1  t 2  t1t 2  1  0, t1  t 2, ,  t 2 t12  t 2  t1t 22  t1  0, ,  t1t 2  1  0  t1t 2  1, , Ans.[3], , Ex.7, , 3, 3, 3, If t1  t 2  t 3 , then points  t1, 2at1  at1  ;  t 2 ,2at 2  at 2  and  t 3 ,2at 3  at 3  are collinear if, , Sol., , (1) t1t 2 t 3  1, (2) t1  t 2  t 3  t1t 2 t 3, Given points will be collinear if, t1, , 2at1  at13, , t2, t3, , 2at 2  at 32 1  0, 2at 3  at 33 1, 1 t1, ,  a 1 t2, 1 t3, ,  t2, t3, , at 32 1  0, at 33 1, , (4) t1  t 2  t 3  1, ,  by C2  2aC1 , , t13, t 32  0, t 33, ,   t1  t 2  t 2  t 3  t3  t1  t1  t 2  t 3   0, ,  t1  t 2  t 3  0, Ex.8, , at13 1, , t1, , 1, , (3) t1  t 2  t 3  0, , [ t1  t2  t3], , Ans.[3], , If a, x1, x 2 are in G.P. with common ratio r, and b, y1,y 2 are in G.P. with common ratio s where s  r  2 , then the, area of the triangle with vertices  a,b  ,  x1,y1  and  x 2 , y 2  is, , , , , , 2, 2, (2) ab  r  s , , 2, (1) ab r  1, , Sol., , 2, (3) ab  s  1, , (4) abrs, , Area of the triangle, , a, 1, , ar, 2, ar 2, , b, bs, bs2, , 1, 1, 1  ab  r  1 s  1 s  r , 2, 1, ,  1,    ab r  1 r  1  ab r 2  1, 2, , , , , , Ans.[1], , 13

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POINT, , EXERCISE # 1, Q.1, , Cartesian coordinates of (2, /4) are, [2] ( 2 , 2), , [1] (2, 2), , [3] ( 2 ,, , 2), , [4] (–, , 2,–, , 2), , Q.2, , The line segment joining the points (3, 4) and (7, 8) is divided by the line 2x + 3y + 7 = 0 in the ratio, [1] 1/5 : 9 externally, [2] 1/5 : 1/9 externally [3] 5 : 1/9 externally, [4] 5 : 9 externally, , Q.3, , If two vertices joining the hypotenuse of a right angled triangle are (0, 0) and (3, 4), then the length of the median, through the vertex having right angle is, [1] 3, [2] 2, [3] 5/2, [4] 7/2, , Q.4, , If (– 4, 6), (2, 3) and (– 2, – 5) are vertices of a triangle, then its incentre is, [1] (– 1, 2), [2] (2, – 1), [3] (1, 2), [4] (2, 1), , Q.5, , Orthocentre of a triangle whose vertex are (8, – 2), (2, – 2) and (8, 6) is, [1] (8, – 2), ,  18 2 , , ,  3 3, , [3] , , [2] (8, 6), , [4] (0, 0), , Q.6, , If P, Q, R are collinear points such that P(7, 7), Q (3, 4) and PR = 10, then R is, [1] (1, 1), [2] (1, – 1), [3] (– 1, 1), [4] (– 1, – 1), , Q.7, , Area of a triangle whose vertices are (a cos, b sin), (– a sin, b cos), and (– a cos, – b sin) is, [1] ab sin cos, , [2] a cos sin, , [3], , 1, ab, 2, , [4] ab, , Q.8, , For what value of k the points (k, 2 – 2k)(1 – k, 2k) and (– 4 –k, 6 – 2k) are collinear ?, [1] 1, – 1/2, [2] 1, 1/2, [3] – 1, 1/2, [4] –1, –1/2, , Q.9, , If the axes are rotated through an angle of 30º in the anti-clockwise direction, then coordinates of points (4 –, 2 3 ) with respect to new axes are, [1] (2,, , Q.10, , 3), , [2] ( 3 , – 5), , [3] (2, 3), , [4] ( 3 , 2), , Reflecting the point (2, – 1) about y-axis, coordinate axes are rotated at 45º angle in negative direction without, shifting the origin. The new coordinates of the points are, , , , 1, , , , 2, , [1]  , , ,, , 3 , , 2, ,  1, , [2] , ,  2, , ,, , 3 , , 2, , , , [3]  , , , , 3, 2, , ,, , 1 , , 2, , [4] none of these, , Q.11, , The locus of a point, which moves in such a way that its distance from the origin (0, 0) is thrice the distance from, x axis is, [1] x2 – 8y2 = 0, [2] 4x2 – y2 = 0, [3] x2 + 8y2 = 0, [4] x2 – 4y2 = 0, , Q.12, , The triangle with vertices 1,5  ;  3,1 and  3, 5  is, [1] isosceles, , [2] equilateral, , [3] right angled, , [4] None of these, 15

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POINT, , Q.13, , , , [1] –1, Q.14, , [2] 1, , [2] x  4 3,y  4 3, , [2] 1: 2, , [2] on x-axis, , [3] 2 : 1, , [4] None of these, , [3] at origin, , [4] None of these, , [2], , 2, , [3] 4, , [4] 2, , 9, 2, , [2] 1, , [3] 3, , [4], , 1, 2, , [4], , 1 1, , a b, , The condition that three points  a,0  ,  0,b  and 1,1 are collinear is, [1], , Q.20, , [4] None of these, , If A  4, 3  ; B  3, 2  and C  2,8  are vertices of a triangle, then the distance of its centroid from y-axis is, [1], , Q.19, , [3] x  4 3,y  4 3, , The distance of orthocentre of the triangle  2,3  ,  4,5  and  1,10  from  2,3  is, [1] 2 2, , Q.18, , [4]  3, , If the vertices of a triangle be  a,b  c  ,  b,c  a  and  c,a  b  , then the centroid of the triangle lies, [1] on y-axis, , Q.17, , 3, , The ratio in which x-axis divides the join of the points  2, 3  and  5,6  is, [1] 2 : 1, , Q.16, , [3], , If the points  4, 4  ,  4,4  and  x,y  form an equilateral triangle, then, [1] x  4 3,y  4 3, , Q.15, , , , If the points 1,1 ,  1, 1 and  3,k are vertices of an equilateral triangle then the value of k will be, , 1 1,  2, a b, , [2], , 1 1,  1, a b, , [3], , 1 1,  0, a b, , If D, E, F are mid points of the sides AB, BC and CA of a triangle formed by the points A  5, 1 ,B  7,6  and, , C 1,3  , then area of DEF is, [1], Q.21, , 2, 5, , [2], , [2] 2, , [3] 4, , [4] 1, , [2] –7, , , , [3] 4, , [4] –4, , , , If the points  0,0  , 2,2 3 and  a,b  be the vertices of an equilateral triangle, then  a,b  , [1]  0, 4 , , Q.24, , [4] 10, , If the area of the triangle with vertices 1,5  ;  4,a  and  2,3  is 3, then a is equal to, [1] 7, , Q.23, , [3] 5, , The distance between feet of perpendiculars drawn from a point  3,4  on both axes is, [1] 5, , Q.22, , 5, 2, , [2]  0,4 , , [3]  4,0 , , [4]  4,0 , , If the point  x, 1 ,  3,y  ,  2,3  and  3, 2  be the vertices of a parallelogram, then, [1] x  2,y  4, , [2] x  1,y  2, , [3] x  4, y  2, , Q.25, , The point on y-axis equidistant from the points  3,2 and  1,3  is, , Q.26, , 3, ,  3, [2]  0,  , [3]  0, , 2, ,  2, If sum of square of distances of a point from axes is 4, then its locus is, [1]  0, 3 , , [1] x  y  2, , [2] x 2  y 2  16, , [3] x  y  4, , [4] None of these, , [4]  0,3 , , [4] x 2  y 2  4, 16

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POINT, , EXERCISE # 2, Q.1, , The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12) the orthocentre of this triangle is, [1] (0, 0), , [2] (0, 24), , [3] (10, 0), ,  13 , ,8 ,  3 , , [4] , , Q.2, , Point P divides the line segment joining A(– 5, 1) and B(3, 5) internally in the ratio  : 1 if Q  (1, 5), R  (7, 2) and, area of PQR = 2, then  equals, [1] 23, [2] 29/5, [3] 31/9, [4] none of these, , Q.3, , Line segment joining (5, 0) and (10 cos, 10 sin) is divided by a point P in ratio 2 : 3. If  varies then locus of P, is a[1] Pair of straight lines [2] Circle, [3] Straight line, [4] Parabola, , Q.4, , The area common to triangle formed by (0, 0), (0, 2p), (2q, 0) and (0, 0), (2q, 0) (2q, 2p) is, [1] 2pq, [2] 4pq, [3] pq, [4] pq/2, , Q.5, Q.6, , ABC is an isosceles triangle. If the coordinates of the base are B(1, 3) and C(– 2, 7) the coordinate of vertex A is, [1] (– 1/2, – 5), [2] (1, 6), [3] (5/6, 6), [4] none of these, An equilateral triangle whose orthocentre is (3, – 2), one side is x-axis then vertex of triangle which is not on xaxis is [1] (3, – 6), [2] (1, – 2), [3] (9, – 2), [4] (3, – 3), , Q.7, , The points (1, 1), (0, sec2), (cosec2, 0) are collinear for [1]  = n, , [2] , , n, 2, , [3]  =, , n, 2, , [4] none of these, , Q.8, , Let A (2, 3) and B (– 4, 5) are two fixed point. A point P moves in such a way that PAB = 12 sq. units then its, locus is, [1] x2 + 6xy + 9y2 + 22x + 66y – 23 = 0, [2] x2 + 6xy + 9y2 + 22x + 66y + 23 = 0, 2, 2, [3] x + 6xy + 9y – 22x – 66y – 23 = 0, [4] none of these, , Q.9, , The locus of the moving point P, sch that 2PA = 3PB where A(0, 0) and B(4, – 3) is, [1] 5x2 + 5y2 + 72x + 54y + 225 = 0, [2] 5x2 – 5y2 + 72x + 54y + 225 = 0, 2, 2, [3] 5x – 5y – 72x + 54y + 225 = 0, [4] 5x2 + 5y2 – 72x + 54y + 225 = 0, , Q.10, , The area of quadrilateral constructed by lines |x| | + |y| = 1 is, [1] 4, [2] 3, [3] 2, , Q.11, , [4] 1, , An equilateral triangle whose circumcentre is (– 2, 5) one side is on y-axis then length of side of triangle is [1] 6, , Q.12, , [2] 2 3, [3] 4 3, [4] 4, The circumcentre of the triangle formed by the points (a cos, a sin), (a cos, a sin), (a cos, a sin) is, [1] (0, 0), ,  a , , a, (cos   cos   cos  ),   (sin   sin   sin  ), , 3,  3 , , , [2] , , [3] (a, 0), [4] None of these, 17

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POINT, , Q.13, , Without changing the direction of coodinates axes, to which point origin should be transferred so that the, equation x2 + y2 – 4x + 6y – 7 = 0 is changed to an equation which contains no term of first degree, [1] (3, 2), , Q.14, , Q.15, , Q.16, , [2] (2, – 3), , [3] (– 2, 3), , [4] none of these, , Two fixed points are A(a, 0) and B(–a, 0) If A – B = , then the locus of point C of triangle ABC will be [1] x2 + y2 + 2xy cot = a2, , [2] x2 – y2 + 2xy tan = a2, , [3] x2 + y2 + 2xy tan = a2, , [4] x2 – y2 + 2xy cot = a2, , Keeping coordinates axes parallel, the origin is shifted to a point (1, – 2), then tranformed equation of x2 + y2 = 2 is, [1] x2 + y2 + 2x – 4y + 3 = 0, , [2] x2 + y2 + 2x + 4y + 3 = 0, , [3] x2 + y2 – 2x – 4y + 3 = 0, , [4] x2 + y2 – 2x + 4y + 3 = 0, , To remove xy term from the second degree equation 5x2 + 8xy + 5y2 + 3x + 2y + 5 = 0, the coordinates axes are, rotated through an angle  then  equals, [1] /2, , Q.17, , [2] /4, , [4] /8, , [3] 3/8, , The point (4, 1) undergoes two successive transformations(i) Reflection about the line y = x, (ii) Translation through a distance 2 units along the positve direction of x-aixs, (iii) The final position of the point is given by the co-ordinates, [1] (4, 3), , Q.18, , Q.19, , [2] (3, 4), , [1] x 2  y 2  2x  2y  0, , [2] x 2  y 2  2x  2y  0, , [3] x 2  y 2  2x  2y  0, , [4] x 2  y 2  2x  2y  0, , If the sum of the distances of a point from the origin and from the line x  2 is always equal to 4, then the locus, of this points is a, [2] circle, , 11, 8, , [2], , 8, 11, , a, , 2, , ac  bd, , , ,  b2 c 2  d2, , , , a, , [2], , 2, , , ,  b2 c 2  d2, , ac  bd, , , , [3], , a, , 2, , , ,  b2 c 2  d2, , , , [4] None of these, , 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, , [1], , Q.24, , [4] None of these, , If the line segment joining the points A  a,b  and B  c,d subtends an angle  at the origin, then cos  is equal, to, [1], , Q.23, , [4] None of these, , [3] 3, , ab  cd, , Q.22, , [3] parabola, , A  6,3  ; B  3,5  ; C  4, 2  and D  x,3x  are four points. If the areas of DBC and ABC are in the ratio 1:2,, then x is equal to, [1], , Q.21, , [4] (1, 4), , The extremities of diagonal of a right angled triangle are  2,0  and  0,2 , then locus of its third vertex is, , [1] straight line, Q.20, , [3] (7/2. 7/2), , xy2, 7, , [2], , xy2, 2, , [3], , xy2, 7, , [4] none of these, , If the vertices of a triangle have integral coordinates, then the triangle is, [1] isosceles, [2] never equilateral, [3] equilateral, , , , [4] None of these, , , , If  a,a  ;  a, a  ; a 3,a 3 are vertices of a triangle then it is, [1] a right angled triangle, , [2] an isosceles triangle, , [3] an equilateral triangle, , [4] None of these, 18

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POINT, , EXERCISE # 3, Q.1, , If a point P is at a equal distance from the points A = (a + b , b –a) and B = (a –b , a + b) , then the locus of P, is, [RPET 93], [1] ax = by, [2] bx = ay, [3] ax + by = 0, [4] bx + ay = 0, , Q.2, , If A  a,0  and B  a,0  are two fixed points and a point P moves such that APB  900 , then locus of P is, [1] x 2  y 2  2a 2, , Q.3, , [4] None of these[RPET, 96], , [2] –5, , [3] 7, , [RPET, 97], [4] –7, , If  8,5  ;  7, 5  and  5,5  are three vertices of a parallelogram, then its fourth vertex is, [1] 10,15 , , Q.5, , [3] x 2  y 2  2a2  0, , If points  a  2, a  4  ;  a,a  1 and  a  4,16  are collinear, then a is equal to, [1] 5, , Q.4, , [2] x 2  y 2  a2, , [2] 15,10 , , [3] 10,5 , , The reflection of the point (4, –3) with respect to the line y = x is, [1] (–4, –3), [2] (3, –4), [3] (–3, 4), , [RPET, 2000], , [4]  5,10 , [RPET 2001], [4] (–3, –4), , Q.6, , 2 ,  1, ,,  are collinear, then x is, If points (x + 1, 2); (1, x + 2) and , x, 1, , x,  1, , , [RPET 2002], , Q.7, , [1] 4, [2] –4, [3] 0, Area of a triangle with vertices (a, b + c); (b, c + a) and (c, a + b) is, [1] abc, [2] a2 + b2 + c2, [3] ab + bc + ca, , [4] none of these, [RPET 2003], [4] 0, , Q.8, , Area of a triangle is 5 and two of its vertices are (2, 1) and (3, – 2). If its third vertex is on the line y = x, + 3, then, it is, [1] ( 1 , 4), , Q.9, , Q.11, , [2] 9x2 – 7y2 + 63 = 0, , [3] 7x2 – 9y2 = 63, , [IIT 83], [4] none of these, , [1] a straight line parallel to x-axis, , [2] a straight line parallel to y-axis, , [3] a circle through origin, , [4] a circle with centre at origin, , Area of a ABC = 20 units and its vertices A and B are (–5, 0) and (3, 0) respectively. If its vertex C lies on the line, x – y = 2, then C is, [IIT 90], [2] (–3, –5), , [3] (–5, 7), , [4] none of these, , , 3  1  1 1 , 1, , The orthocentre of triangle with vertices  2, 2  ,  2 ,  2  and  2,   is, 2, , , , , 3, 3 3, [1]  2 ,  6 , , , , Q.13, , [4] (13/2 , 19/2), , If P  (1, 0); Q(–1, 0) and R  (2, 0) are three given points, then locus of a point S satisfying the relation, SQ2 + SR2 = 2SP2 is, [IIT 88], , [1] (3, 5), , Q.12, , [3] ( 7/2, 13/2), , If A  (0, 4); B  (0, –4) and |AP – BP| = 6 then the locus of P is, [1] 9x2 – 7y2 = 63, , Q.10, , [2] ( 10 , 13), , 1, , [2]  2 ,  , 2, , , [IIT 93], , 5 3 2, [4]  4 , 4 , , , ,  1 1, [3]   , ,  2 2, , If P 1,2  ; Q  4,6  ; R  5,7  and S  a,b  are vertices of a parallelogram PQRS, then, [1] a  2,b  4, , [2] a  3,b  4, , [3] a  2,b  3, , [IIT, 98], , [4] a  3,b  5, , 19

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POINT, , Q.14, , If x1, x2, x3 and y1, y2, y3 are in GP’s whose common ratios are equal, then points (x1, y1); (x2, y2) and (x3, y3) are, [IIT 99, AIEEE 2003], [1] collinear, [2] on a circle, [3] vertices of a triangle [4] none of these, , Q.15, , The incentre of the triangle with vertices (1, 3 ), (0, 0) and (2, 0) is, [2] (2 / 3, 1/ 3 ), , [1] (1, 3 / 2), Q.16, , Q.17, , [IIT 2000, AIEEE 2002], , [3] (2 / 3, 3 / 2), , [4] (1, 1/ 3 ), , Orthocentre of triangle with vertices ( 0 , 0 ) , ( 3, 4) and ( 4 , 0) is, [1] ( 3 , 5/4), [2] ( 3 , 12), [3] ( 3 , 3/4), , [IIT 2003], [4] (3, 9), , Locus of centroid of the triangle whose vertices are  acos t, a sin t  ,  b sin t,  bcos t  and 1,0  , where t is a, parameter, is, [1]  3x  1   3y   a2  b2, , [2]  3x  1   3y   a 2  b2, , [3]  3x  1   3y   a 2  b2, , [4]  3x  1   3y   a 2  b 2, , 2, , 2, , Q.18, , [AIEEE, 03], 2, , 2, , 2, , 2, , 2, , 2, , If the equation of the locus of a point equidistant from points  a1,b1  and  a2 ,b2  is  a1  a2  x   b1  b2  y  c  0 ,, then c is equal to, , [AIEEE, 03], , Q.19, , 1 2, 1 2, a2  b22  a12  b12, a1  a22  b12  b22, [3] a12  a 22  b12  b 22, [4], 2, 2, A(2 , –3) and B(–2,1) are vertices of a ABC, if centroid of the triangle lies on the line 2x + 3y = 1, then locus of, the vertex C is, [AIEEE 2004], [1] 3x + 2y = 5, [2] 2x – 3y = 7, [3] 2x + 3y = 9, [4] 3x – 2y = 3, , Q.20, , If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (-1, 2) and (3, 2), then the, , [1], , a12  b12  a22  b22, , , , [2], , , , , , , , centroid of the triangle is, , 1 7, 3 3, , [1]  , , Q.21, , [AIEEE-2005], ,  7,  3, , [2] 1, , ,  1 7 , , ,  3 3, , [3] , , , , , 7, 3, , [4]  1, , , Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of, the triangle is 1, then the set of values which k can take is given by, [1] {–1, 3}, , Q.22, , [2] {–3, –2}, , [3] {1, 3}, , [AIEEE-2007], [4] {0, 2}, , Let O(0, 0), P(3, 4), Q(6, 0) be vertices of a OPQ. R is a point inside this triangle such that areas of ’s OPR,, PQR, OQR are equal. Then R is equal to :, [1] (4/3, 3), , Q.23, , [2] (3, 2/3), , [IIT-JEE -2007], [3] (3, 4/3), , [4] (4/3, 2/3), , Consider three points, P = (–sin( – ), – cos),, Q = (cos ( – ), sin ), and, , R = (cos ( – ), sin ( – )), , where 0 <  < /5. Then :, , Q.24, , [IIT-JEE -2008], , [1] P lies on the line segment RQ, , [2] Q lies on the line segment PR, , [3] R lies on the line segment QP, , [4] P, Q, R are non-collinear, , Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the, distance of any one of them from the point (1, 0) to the distance from the point (–1, 0) is equal to 1/3. Then the, circumcentre of the triangle ABC is at the point :, [1] (5/4, 0), , [2] (5/2, 0), , [AIEEE -2009], [3] (5/3, 0), , [4] (0, 0), , 20

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POINT, , Q.25, , If the line 2 x  y  k passes through the point which divides the line segment joining the points (1, 1) and, (2, 4) in the ratio 3 : 2, then k equals :, (1), , Q.26, , 29, 5, , [AIEEE – 2012], , (2) 5, , (3) 6, , (4), , 11, 5, , The x – coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0,1)(1,1), and (1,0) is :, , [JEE Mains – 2013], (2) 1  2, , (1) 1  2, , (3) 2  2, , (4) 2  2, , ANSWER KEY, EXERCISE - 1, Que., Ans., , 1, 3, , 2, 4, , 3, 3, , 4, 1, , 5, 1, , 6, 3, , 7, 4, , 8, 3, , 9, 2, , 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25, 1 1 3 3 3 2 2 1 3 2 2 1 2 3 1 2, , 26, , 4, , EXERCISE - 2, Que ., Ans., , 1, 1, , 2, 1, , 3, 2, , 4, 3, , 5, 3, , 6, 1, , 7, 2, , 8, 3, , 9, 4, , 10, 3, , 11, 3, , 12, 1, , 13, 2, , 14, 4, , 15, 1, , 16, 2, , 17, 2, , 18, 1, , 19, 3, , 20, 1, , 21, 2, , 22, 1, , 23, 2, , 24, 3, , EXERCISE - 3, , 21