A line `x-y+1=0` cuts the y-axis at `A.` This line is rotated about `A` in the clockwise direction through `75^@.` Find the equation of the line in the new position
(A) `sqrt(3)y + x = sqrt3` (B) `sqrt x + y = sqrt3` (C) `x + sqrty = 1` (D) `sqrt x + y = 1`

If the pair of lines sqrt(3)x^2-4x y+sqrt(3)y^2=0 is rotated about the origin by pi/6 in the anticlockwise sense, then find the equation of the pair in the new position.

If the pair of lines sqrt(3)x^2-4x y+sqrt(3)y^2=0 is rotated about the origin by pi/6 in the anticlockwise sense, then find the equation of the pair in the new position.

Find the equation of straight lines through the point A(3,-2) and inclined at 60^(@) to the line sqrt(3)x+y=1 .

Find the angle between the lines y-sqrt(3)x-5=0 and sqrt(3)y-x+6=0 .

The equation of the circumcircle of the triangle formed by the lines y + sqrt(3)x = 6, y - sqrt(3) x = 6 and y = 0 , is-

The graph of straight line y = sqrt(3)x + 2sqrt(3) is :

If the represented by the equation 3y^2-x^2+2sqrt(3)x-3=0 are rotated about the point (sqrt(3),0) through an angle of 15^0 , on in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is (1) y^2-x^2+2sqrt(3)x+3=0 (2) y^2-x^2+2sqrt(3)x-3=0 (3) y^2-x^2-2sqrt(3)x+3=0 (4) y^2-x^2+3=0

Find the angle between the lines y=(2-sqrt3)(x+5) and y=(2+sqrt3)(x-7) .

Find the angles between the pairs of straight lines: x+sqrt(3)y-5=0\ a n d\ sqrt(3)x+y-7=0

(v) Solve for x and y if (sqrt2 x+sqrt 3 y)=0 and (sqrt3 x- sqrt8 y)=0