A straight line L=0  passes through  the point 5,−4 is inclined at an angle of 60° to the line 3x+y−3=0,then the equation of the lineL=0 is

The equation of any line which makes an angle α with the line ax+by+c=0 and passing  through the point x1,y1 is y−y1=tan⁡(θ±α)x−x1 here tan⁡θ=−ab Here tan⁡θ=−31⇒θ=120∘ and α=60∘ Equations of the required lines are y+4=tan⁡120∘±60∘(x−5)It implies that y+4=tan180°x−5y+4=0Ory+4=tan60°x−5y+4=3x−5 Therefore, the equations of the required lines are y+4=0,3x−y−53−4=0