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.