The slope of the line #AB# given by #y=1/2x# is #1/2#
Let the slope of other line be #m# i.e. it is of the type #y=mx+c#. But as it passes through #(1,1/2)#, we have #1/2=m+c# or #c=1/2-m#.
As the slope of two lines is #1/2# and #m#, the angle between them will be given by
#tan(+-A)=(m-1/2)/(1+mxx1/2)=(2m-1)/(2+m)#
As angle is #pi/6# and #tan(pi/6)=+-1/sqrt3# and hence
either #(2m-1)/(2+m)=1/sqrt3# i.e. #2sqrt3m-sqrt3=2+m# and #m=(2+sqrt3)/(2sqrt3-1)#
and then #c=1/2-(2+sqrt3)/(2sqrt3-1)=(2sqrt3-1-4-2sqrt3)/(4sqrt3-2)=-5/(4sqrt3-2)#
and equation of line is #y=(2+sqrt3)/(2sqrt3-1)x-5/(4sqrt3-2)#
or #(4sqrt3-2)y-(4+2sqrt3)x+5=0#
or #(2m-1)/(2+m)=-1/sqrt3# i.e. #2sqrt3m-sqrt3=-m-2# and #m=-(2-sqrt3)/(2sqrt3+1)#
and then #c=1/2+(2-sqrt3)/(2sqrt3+1)=(2sqrt3+1+4-2sqrt3)/(4sqrt3+2)=5/(4sqrt3+2)#
and equation of line is #y=-(2-sqrt3)/(2sqrt3+1)x+5/(4sqrt3+2)#
or #(4sqrt3+2)y+(4-2sqrt3)x-5=0#
#((4sqrt3-2)y-(4+2sqrt3)x+5)((4sqrt3+2)y+(4-2sqrt3)x-5)(2y-x)=0#
graph{((4sqrt3-2)y-(4+2sqrt3)x+5)((4sqrt3+2)y+(4-2sqrt3)x-5)(2y-x)=0 [-4, 4, -2, 2]}
