If perpendicular distance from the point (3, 1) to the line (2a + b)x + (a + b )y - b = 0 (where a, b are not 0) maximum then (b-a)/(b+a) is equal to: 4 9(b) 10 (c) 11 (d) 12

Locus of the image of a point A(0,0) in the line (2a+b)x+(3a-2b)y+(b-5a)=0 is (where a and b are parameters).

Find the perpendicular distance of the point A(1,0,1) to the line through the points B(2,3,4) and C(-1,1,-2)

Find the perpendicular distance of the point A(1,0,1) to the line through the points B(2,3,4) and C(-1,1,-2)

If the length of the perpendicular from the point (1,1) to the line a x-b y+c=0 be unity, show that 1/c+1/a-1/b=c/(2a b)

The lines ax + by + c = 0, where 3a + 2b + 4c= 0, are concurrent at the point (a) (1/2 ,3/4) (b) (1,3) (c) (3,1) (d) (3/4 ,1/2)

The lines ax + by + c = 0, where 3a + 2b + 4c= 0, are concurrent at the point (a) (1/2 ,3/4) (b) (1,3) (c) (3,1) (d) (3/4 ,1/2)

The sum of the minimum distance and the maximum distance from the point (4, -3) to the circle x^(2) + y^(2) + 4x - 10y - 7 = 0 is a)20 b)12 c)10 d)16

a,b,cgt0 and a+b+c=12 . If a^(2)b^(4)c^(3) is maximum then (4a)/(b) is

Let A B C be a triangle. Let A be the point (1,2),y=x be the perpendicular bisector of A B , and x-2y+1=0 be the angle bisector of /_C . If the equation of B C is given by a x+b y-5=0 , then the value of a+b is (a) 1 (b) 2 (c) 3 (d) 4