Line rotated by an angle ±θ from a given line whose slope is m_{b}

The slope of a given line is: m_{b} = tan θ_{b}
or θ_{b} = tan^{1} m_{b}
The angles of the lines which are inclined by an angle θ from the original line are: θ_{1} = θ_{b} − θ
and θ_{2} = θ_{b} + θ
Hence the slopes of the lines are: 
m_{1} = tan θ_{1} = tan (θ_{b} − θ) 
m_{2} = tan θ_{2} = tan (θ_{b} + θ) 
And the inclined lines equations which are passing throught the point (x_{0} , y_{0}) are:
y − y_{0} = m_{1}(x − x_{0}) = tan(θ_{b} − θ)(x − x_{0}) 
y − y_{0} = m_{2}(x − x_{0}) = tan(θ_{b} + θ)(x − x_{0}) 
Or expressed by m_{b}
y − y_{0} = tan(tan^{1} m_{b} ± θ)(x − x_{0}) = tan(θ_{b} ± θ)(x − x_{0}) 
the plus and minus signs stands for the two lines equations which lies either side of the original line.

Example: find the equation of a line that is inclines by 15 degrees to the right of the line
2x −y + 3 = 0 and is passing through the point (1 , 5).

The slope of the given line is: 
m_{b} = 2 
The angle of the given line is: 
θ_{b} = tan^{1} 2 = 63.43 degree 
The angle of right incline line is: 
θ_{1} = 63.43 − 15 = 48.43 degree 
The slope of right incline line is: 
m_{θ} = tan 48.43 = 1.13 
Hence the line equation is:
(y − 5) = tan(63.43 − 15)(x − 1) 
(y − 5) = 1.13(x − 1) 
1.13x − y + 3.87 = 0 
