Point & Straight Line (Point)

Point & Straight Line POINT RECTANGULAR CARTESIAN CO-ORDINATE SYSTEMS DISTANCE FORMULA The distance between two points P x1,y1 and Q x2 ,y 2 is given by We shall right now focus on two-dimensional co-ordinate geometry in which two perpendicular Note : lines called co-ordinate axes (x-axis and y-axis) are (i) In particular the Distance of a point P x, y from the used to locate a point in the plane. origin (ii) Distance between two polar co-ordinates A(r 1 ,  1 ) and B(r 2 ,  2 ) is given by Ex.1 The distance between [1] 115 [2] 109 and [3] 91 is [4] 11 O is called origin. Any point P in this plane can be Sol. represented by a unique ordered pair (x, y), which Ans. [2] are called co-ordinates of that point. x is called x and is co-ordinate or abscissa and y is called y co-ordinate Ex.2 The distance between or ordinate. The two perpendicular lines xox and [1] 3 [2] 1/2 [3] 7 [4] 5 yoy divide the plane in four regions which are called    2 2 quadrants, numbered as shown in the figure. Sol. PQ  2  3  2.2.3cos   6  6    Let us look at some of the formulae linked with points 1    13  12  cos     13  12   7 Ans. [3] now. 3 2   1 Point & Straight Line Ex.3 Find the value of x, if the distance between the points (ii) For given four points: (x, –1) and (3, 2) is 5  Sol. Let P(x ,–1) and Q(3, 2) be the given points. Then PQ = 5 (given) 2 ( x  3)  ( 1  2)  (x 2 – 3)2 + 9 = 25 =5  rhombus  is a rectangle   x = 7 or x = – 1 AB  CD, BC  DA, AC  BD  ABCD is a parallelogram APPLICATION OF DISTANCE QUADRILATERAL DIAGONALSANGLE FORMULAE BETWEEN DIAGONALS (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: (i) Parallelogram Not equal (ii) Rectangle Equal  Collinear: If the sum of any two distances is equal to the third. (iii) Rhombus Not equal  Vertices of an equilateral triangle if AB = BC = CA (iv) Square Equal  Vertices of an isosceles triangle if AB = BC or BC Note : = CA or CA = AB (i) Diagonal of square, rhombus, rectangle and Vertices of a right angled triangle parallelogram always bisect each other. etc., (ii) Diagonal of rhombus and square bisect each other at right angle.  Ex.4 If and C 5,2 are three points then A, B, C are (iii) Four given points are collinear, if area of quadrilateral is zero. [1] collinear [2] vertices of an equilateral triangle [3] vertices of right angled triangle Ex.5 Points [4] none of these and are [1] collinear [2] vertices of equilateral triangle [3] vertices of isoscele triangle Sol. [4] none of these 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 are vertices of a right angled triangle. Ans. [3] Sol. Clearly Ans. [1] . Hence A,B,C are collinear.. 2 Point & Straight Line (ii) For external division 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) For internal division x= 3  4  2  6 32 or and y = 35  23 32 x = –24 and y = 9 So the co-ordinates of P are (–24, 9) (2) For external division Ex.7 Find the co-ordinates of points which trisect the line segment joining (1, – 2) and (– 3, 4). (3) Co-ordinates of mid point of PQ Put (4) Co-ordinates of any point on the line segment joining two points P(x1, y1) and Q x2 ,y 2 are 1. DIVISION BY AXES: PQ is divided by Then AP = PQ = QB =  (say)  PB = PQ + QB = 2 and AQ = AP + PQ = 2 AP : PB =  : 2 = 1 : 2 and AQ : QB = 2 :  =2:1  (i) x-axis in the ratio 2 (ii) y-axis in the ratio So P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1 2 2. Sol. Let A (1, –2) and B(–3, 4) be the given points. Let the points of trisection be P and Q. DIVISION BY A LINE: A line  divides PQ in the ratio the co-ordinates of P are  1 3  2  1 1 4  2  2  ,   1 2 1 2   1   or   3 , 0    and the co-ordinates of Q are Ex.6 Find the co-ordinates of the point which divides the line segment joining the points (6, 3) and (– 4, 5) in the ratio 3 : 2 (i) internally and (ii) externally. Sol. Let P (x, y) be the required point.  2  3  1 1 2  4  1 ( 2)  ,   2 1 2 1    5  or   3 , 2    1  3   Hence, the points of trisection are   , 0  and   5    , 2 .  3  (i) For internal division : Ex.8 The co-ordinates of point of internal and external x= 3  4  2  6 32 x=0 and and y = y= 35  23 32 21 5  So the co-ordinates of P are  0,  21   5  division of the line segment joining two points or and 3,4 in the ratio 2 : 3 are respectively [1] (2, 3) (11, 3) [3] (1, 3) (–11, 3) [2] (3, 1) (3, –11) [4] (1, –3) (11, –3) 3 Point & Straight Line Centroid, Incentre & Excentre : Sol. Internal division If A (x1, y1), B(x2, y2), C(x3, y3) are the vertices of triangle ABC, whose sides BC, CA, AB are of lengths a, b, c respectively, then the co-ordinates of the special points of triangle ABC are as follows : Hence point 3,1 External division x 1  x 2  x 3 y1  y 2  y 3  ,  3 3    Centroid G   Hence point  ax 1 bx 2  cx 3 ay 1 by 2  cy 3 , a b c  abc Ans.[2] Incentre I   Ex.9 The ratio in which the line divides the line segment joining the points 1,2 and [1] 4 9 9 [2] 4 [3] 1 3 is   ,and  Excentre (to A) I1  ax1  bx 2  cx 3  ay1  by 2  cy 3    a  b  c ,  a  b  c  and so on.     3 [4] 4 Notes : (i) Incentre divides the angle bisectors in the ratio, ax 1  by1  c Sol. Required ratio =  ax  by  c (b + c) : a; (c + a) : b & (a + b) : c. 2 2 (ii) Incentre and excentre are harmonic conjugate of Ans. [1] each other w.r.t. the angle bisector on which they lie. Ex.10 The points of trisection of line joining the points (iii) Orthocentre, Centroid & Circumcentre are always A 2,1 and B 5,3 are collinear & centroid divides the line joining orthocentre & circumcentre in the ratio 2 : 1. (iv) In an isosceles triangle G, O,  & C lie on the same [1] [2] line and in an equilateral triangle, all these four points coincide. [3] [4] (v) In a right angled triangle orthocentre is at right angled vertex and circumcentre is mid point of hypotenuse (vi) In case of an obtuse angled triangle circumcentre Sol. A P1 ' P2 ' and orthocentre both are out side the triangle. Ans.[1] Ex.11 The ratio in which the lines joining the and [2] 2 : 3 y 1 Sol. Required ratio =  y 2  x1  x 2  x3 y1  y 2  y3  ,  3 3   (x1, y1), (x2, y2) (x3, y3) are  divided by x-axis is [1] 3 : 2 Ex.12 Find the co-ordinates of (i) centroid (ii) in-centre of the triangle whose vertices are (0, 6), (8, 12) and (8, 0). Sol. (i) We know that the co-ordinates of the centroid of a triangle whose angular points are [3] 4 : 3 [3] 3 : 4 Ans.[2] So the co-ordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0) are  0  8  8 6  12  0  ,   3 3   16   or  3 , 6  .   4 Point & Straight Line (ii) Let A (0, 6), B (8, 12) and C(8, 0) be the vertices Ex.15 If 1,4 is the centroid of a triangle and its two of triangle ABC. vertices are and then third vertices is Then c = AB = (0  8)2  (6  12)2 = 10, [1] (7, 8) [2] (8, 8) [3] (8, 7) [4] (6, 8) Sol. Let the third vertices of triangle be x,y then b = CA = (0  8 )2  (6  0 )2 = 10 and a = BC = (8  8)2  (12  0)2 = 12. So third vertex is (8, 8). The co-ordinates of the in-centre are Ans.[2]  ax 1  bx 2  cx 3 ay 1  by 2  cy 3   ,  abc abc   or Ex.16 If 0,1 , 1,1 and 1,0 are middle points of the  12  0  10  8  10  8 12  6  10  12  10  0  ,   12  10  10 12  10  10    160 192  ,    32 32  sides of a triangle, then find its incentre is or or (5, 6) [1] 2  2 , 2  2 [2] 2  2,  2  2 [3] 2  2 , 2  2 [4] 2  2 ,  2  2 Sol. Let A x1, y1 , B x2 , y 2 and C x 3 , y3 are vertices Ex.13 Centroid of the triangle whose vertices are of a triangle, then 0,0 , 2,5 and 7, 4 is [1] (4, 3) [2] (3, 4) [3] (3, 3) Sol. [4] (3, 5) Solving these equations, we get A 0,0 , B 0,2 and Ans.[3] C 2,0 Now Ex.14 Incentre of triangle whose vertices are Thus incentre of is Ans.[1] is [1] (1, 1) [2] (0, –1) [3] (–1, 0) [4] (1, 0) Sol. Using distance formula  Ex. 17 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) (2) (–4, 7) (3) (–4, –7) (4) ( 4 , 7) Sol. Let C (  , ) be the third vertex , Ans.[3] 5 –  = – 13 .... (1) 2 –3 = 13 ..... (2) Solving (1) and (2) , ( , ) = ( –4 , –7) Ans.[3] 5 Point & Straight Line Ex.18 If O 0,0 ; A 3,0 and B 0, 4 are vertices of a (iii) When two vertices be on x-axis say a,0 , b,0 and triangle, then its circumcentre is [1] (1, 1) 3  [2]  2 , 2    third vertex be h,k , then its area  4  3 [3] 1, 3  [4]  2, 2      (iv) When two vertices be on y-axis say 0,c , 0,d and third vertex be h,k , then its area Sol. Let it be P x,y . Then (v) Area of the triangle formed by coordinate axes and the line is c2 2ab (vi) When ABC is right angled triangle and circumcentre , then Ans.[2] (vii) When ABC is equilateral triangle, then AREA OF TRIANGLE AND QUADRILATERAL 1. (viii) When D, E, F are the mid points of the sides AB, BC, CA of the triangle ABC, then its area AREA OF TRIANGLE (Cartesian Coordinates) Let A x1, y1 , B x2 , y 2 and C x 3 , y3 are vertices Note: of a triangle, then x1 y1 1 x3 y3 1 Area of a triangle is always taken to be non-negative. So always use mod sign while using area formula. Area of triangle Ex.19 If the vertices of a triangle are and 3,5 then its area is Condition of collinearity: [1] Three points A x1, y1 , B x 2 , y 2 , C x3 ,y 3 collinear if the area of if x1 y1 1 x3 y3 1 are [3] 5 sq. unit i.e., [2] 12 sq. unit [4] 25 sq. unit Sol. Ans.[1] Particular cases: 2. (i) When one vertex is origin i.e., if the vertices are 0,0 ; x1, y1 25 sq. unit 2 and x2, y2 then its area AREA OF QUADRILATERAL If x1, y1 , x 2 ,y 2 , x 3 ,y 3 and x 4 ,y 4 are vertices of a quadrilateral then its area then its area is zero. 6 Point & Straight Line Area of PAB = 10 Now, Note: (i) If the area of quadrilateral joining four points is zero then those four points are collinear. 1  2 (ii) If two opposite vertex of rectangle are x1, y1 and x 2 , y 2 then its area may be (iii) If two opposite vertex of a square are A x1, y1 and C x 2 , y2 then its area is Ex.20 If x y 1 3 4 1 = ± 10 5 2 1  6x + 2y – 26 = ± 20  6x + 2y – 46 = 0 or 6x + 2y – 6 = 0  3x + y – 23 = 0 or 3x + y – 3 = 0 Solving x – 3 y – 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. Thus, the co-ordinates of P are (7, 2) or (1, 0) and are vertices of a 3. AREA OF A TRIANGLE (Polar Coordinates) If quadrilateral then its area is and r , are vertices of a triangle then its area . 41 [1] sq. units 2 [2] 41 sq. units [3] 20 sq. units [4] 22 sq. units 1 Sol.  1(4)  3(1)  3 2  5 4  5 7  4 2  4 1  1 7  2 Ex.23 The area of a triangle with vertices  5 7  4 2  4 1  1 7  sq. units Ans.[1] [1] 5 2 2 a 4 [2] 2 5 2 a 4 [3] 5 3 2 a 4 [4] 2 3 2 a 4 Ex.21 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 [3] (a – b)2 [4] 2(a – b)2 Sol. We know that area of square Ex.22 If the co-ordinates of two points A and B are (3, 4) and (5, –2) respectively. Find the co-ordinates of any point P if PA = PB and Area of PAB = 10. Sol. Let the co-ordinates of P be (x, y). Then PA = PB  PA2 = PB2  (x – 3)2 + (y – 4)2 = (x – 5)2 + (y + 2)2 x – 3y – 1 = 0 Ans.[3] 2 Ans.[3]  Sol. 4. AREA OF ATRIANGLE WHEN EQUATIONS OF ITS SIDES ARE GIVEN: If are sides of a triangle then its area is given by 1 a1 a2 b1 b2 c1 c2 a3 b3 c3 2 when C1, C2, C3 are cofactors of c1,c 2 ,c 3 in the determinant. 7 Point & Straight Line Ex.24 If x – y = 1, x + 2y = 0 and 2x + y = 3 are sides of a triangle, then its area is 2 [1] 3 [2] 3 2 [3] 2 [4] 1 2 2 1 1 1 1 1 2 Sol. Area = 2(3 )(3)3 1 2 0  54  36  3 . 2 1 3 Ex.25 The locus of a point such that the sum of its distances from the points 0,2 and is 6 is [1] 9x2 + 5y2 = 45 [2] 5x2 + 9y2 = 45 [3] 4x2 + 7y2 = 35 Ans.[1] [4] 9x2 + 5y2 = 50 LOCUS OF A POINT The locus of a moving point is the path traced out by Sol. Let P h,k be any point on the locus and let A 0,2 that point under one or more given conditions. and be the given points. How to find the locus of a point : Let (x1, y1) be the co-ordinate of the moving points say P. Now By the given condition 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 where then the locus of P is a circle. (iii) Locus of P if A and B are fixed. (a) Circle if (b) Circle with diameter AB if (c) Ellipse if (d) Hyperbola if Hence, locus of h,k is Ans.[1] 8