Point on `x-y+1=0` Which lies at minimum distance from the point `(1,4)` is
(a)(2,1)
(b)(3,4)
(c)(2,3)
(d)(5,6)

Find the point on the curve y^2=2x which is at a minimum distance from the point (1,\ 4) .

The minimum distance of the point (1, 2, 3) from x-axis is

The distance of the point (-3,4) from the x-axis is : (a) 3 (b) -3 (c) 4 (d) 5

Plot the points A(4,1),B(-3,2),C(2,-3),D(-4,1),E(-4,-4) .

The distance of the point (-3,4) from the x-axis is : a) '3 (b) -3 (c) 4 (d) 5

The distance of the point (-3,4) from the x-axis is : 3 (b) -3 (c) 4 (d) 5

The point on the line (x-2)/1=(y+3)/(-2)=(z+5)/(-2) at a distance of 6 from the point (2,-3,-5) is (a). (3,-5,-3) (b). (4,-7,-9) (c). 0,2,-1 (d). none of these

A plane 2x+3y+5z=1 has a point P which is at minimum distance from line joining A(1, 0, -3), B(1, -5, 7), then distance AP is equal to

A plane parallel to y -axis passing through line of intersection of planes x+y+z=1 & 2x+3y-z-4=0 which of the point lie on the plane (a) (3,2,1) (b) (-3,0,1) (c) (-3,1,1) (d) (3,1,-1)

The point on the circle (x-3)^2 + (y-4)^2 = 4 which is at least distance from the circle x^2 + y^2 = 1 is : (A) (3/5, 4/5) (B) (9/5, 12/5) (C) (9, 12) (D) none of these