Distance possible to draw a line which belongs to all the given family of lines `y-2x+1+lambda_1(2y-x-1)=0,3y-x-6+lambda_2(y-3x+6)=0,a x+y-2+lambda_3(6x+a y-a)=0` , then (a)`a=4` (b) `a=3` (c)`a=-2` (d) `a=2`

Length of the shortest chord of the parabola y^(2)=4x+8, which belongs to the family of lines (1+lambda)y+(lambda-1)x+2(1-lambda)=0 is

The equation of line belonging to the family of lines (5x+3y-2)+lambda(3x-y-4)=0 , lambda in R ,and at greatest distance from (0,0) is

The equation of line belonging to the family of lines (5x+3y-2) + lambda (3x-y-4)=0, lambda in R , and at greatest distance from (0,2) is (A) x-y-2=0 (B) x+y=0 (C) x-y=0 (D) 2x-y-3=0

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0

The value of the lambda if the lines (2+3y+4)+lambda(6x-y+12)=0 are

Locus of the image of the point (1,1) in the line (5x-2y-7)+lambda(2x-3y+6)=0lambda in R, is

Locus of the image of the point (1,1) in the line (5x-2y-7)+lambda(2x-3y+6)=0 (lambda in R'), is

The centre of family of circles cutting the family of circles x^(2)+y^(2)+4x(lambda-(3)/(2))+3y(lambda-(4)/(3))-6(lambda+2)=0 orthogonally,lies on

If the lines x - 2y - 6 = 0 , 3x + y - 4 and lambda x +4y + lambda^(2) = 0 are concurrent , then