I have the equation of a beam that looks like this: $$(x + y - 5) + k(2x - 3y) = 0$$
I know that the angular coefficient of a $60^\circ$ angle is equivalent to the root of 3.
$$m = \sqrt3$$
Though, how do I determine k so that the line of the beam is parallel a 60 degrees angle with the x-axis?
The given equation is $(2k+1)x+(1-3k)y-5=0$. The slope of the beam is ${2k+1\over 3k-1}$ which must be equal to $\sqrt 3$. That is $${2k+1\over 3k-1}=\sqrt 3$$ Can you continue and find $k$?