A vertex of an equilateral triangle is at (2, 3), and th equation of the opposite side is `x+y=2`, then the equaiton of the other two sides are (A) `y=(2+sqrt(3)) (x-2), y-3=2sqrt(3)(x-2)` (B) `y-3=(2+sqrt(3) (x-2), y-3= (2-sqrt(3) (x-2)` (C) `y+3=(2-sqrt(3)(x-2), y-3=(2-sqrt(3) (x+2)` (D) none of these

Number of equilateral triangle with y=sqrt3(x-1)+2;y=- sqrt3x as two of its sides is

The equation of the tangents to the ellipse 4x^2 + 3y^2 = 5 , which are inclined at 60^0 to the X-axis are : (A) y = x/sqrt(3) +- sqrt(65/12) (B) y = sqrt(3) +- sqrt(65/12) (C) y = sqrt(3)x +- sqrt(12/65) (D) none of these

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

sqrt (2x) -sqrt (3y) = 0sqrt (3x) -sqrt (3y) = 0

If the equation of base of an equilateral triangle is 2x-y=1 and the vertex is (-1,2), then the length of the sides of the triangle is sqrt((20)/(3)) (b) (2)/(sqrt(15))sqrt((8)/(15)) (d) sqrt((15)/(2))

A circle C of radius 1 is inscribed in an equilateral triangle PQR . The points of contact of C with the sides PQ, QR, RP and D, E, F respectively. The line PQ is given by the equation sqrt(3) +y-6=0 and the point D is ((sqrt(3))/(2), 3/2) . Equation of the sides QR, RP are : (A) y=2/sqrt(3) x + 1, y = 2/sqrt(3) x -1 (B) y= 1/sqrt(3) x, y=0 (C) y= sqrt(3)/2 x + 1, y = sqrt(3)/2 x-1 (D) y=sqrt(3)x, y=0

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

Evaluate : ( sqrt(2)x + sqrt(3)y ) ( sqrt(2)x + sqrt(3)y )

The equation of the base of an equilateral triangle ABC is x+y=2 and the vertex is (2,-1). The area of the triangle ABC is: (sqrt(2))/(6) (b) (sqrt(3))/(6) (c) (sqrt(3))/(8) (d) None of these