See Fig. Let ∠XAB = α so that

tanα = 1 -0/3 - 2 = 1 or α = 45°

Since ∠XAC = α + 15° = 60°, the equation of the line AC (point C is the new position of point B) is

y - 0 tan 60°(x - 2) = √3(x -2)

Therefore, equation of the line AB in its new position is

√3x - y - 2√3 = 0

The value of x = 2 − 1/ √2 gives the position of point B, when (bar)AB is rotated about point A through angle 15° in clockwise sense.