By Rakesh, Gupta

ISBN-10: 8182472253

ISBN-13: 9788182472259

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**Example text**

Find the equation of Perpendicular bisector of the line segment AB having end points A (3,5) and B (6,8). We know that perpendicular bisector of a line segment AB is the locus of a moving point whose distance from the end points A and B are always equal. 8)2 (X-3)2 +(y-5)2 = (X-6)2 + (y -8)2 x2 +y2-6x -lOy +34 = x2 +y2 -12x -16y +100 6x +6y -66 = 0 x +y -11 = 0, is the required locus of point P. Find the locus of a point P such that PA 4 2PA2 + PB2 = 3 (ii) PB 3 where co-ordinates of points A and Bare (-1,2) and (3,4) respectively.

1,2) b. 1) c. (0,3) d. Both (b) and (c) The locus of a point which lies on the line joining the point (3A) and (2,2) is c. 2x = y-2 d. x = 2y- 2 a. x = 2y + 2 b. 2x = Y + 2 /42/ 63. The locus of a point P moving on a plane such that PA = PB; where A(4,1) and B(3,5) is c. 2x +8y -17 = 0 d. x +8y +17 = 0 a. 2x -8y +17 = 0 b. x -8y +17 = 0 64. If A(1,3) and B(5,7) are two fixed points, then the locus of point P(x, y) for which 2PA2 +3PB2 = 10 is . b. 5x 2+ y2-34x -54y +232 = 0 a. x2 +5y2 -34x -54y +232 = 0 2 c.

5x 2+ y2-34x -54y +232 = 0 a. x2 +5y2 -34x -54y +232 = 0 2 c. 5x +5 y 2_34x -54y +232 = 0 d. None of these. 65. For the points A(a, 0) and B(-a, 0) the locus of a point C(x, y) such that LACB is right Angle is given by a. x2_y2 = a 2 b. _x 2 + y2 =a2 c. x2+ y2 = _a 2 d. x2 + y2 = a 2 66. x2 y2 The point lying on the curve ~ - 11 = 1 is a. (a secS , b tanS) b. (a sinS, b cosS) 67. 68. c. (a sinS, -b cose) d. None of these. x2 y2 The point lying on the curve 2"" + -2 = 1 a b b. (a sine, b cosS) a.

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