Given: point \[\left( \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{3} \right)\] and \[\mathbf{\theta }\text{ }=\text{ }\mathbf{75}{}^\circ \]
Condition of line: \[\left( \mathbf{y}\text{ }\text{ }\mathbf{-y1} \right)\text{ }=\text{ }\mathbf{m}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{-x1} \right)\]
where, \[\mathbf{m}\text{ }=\text{ }\mathbf{incline}\text{ }\mathbf{of}\text{ }\mathbf{line}\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{\theta }\text{ }\mathbf{and}\text{ }\left( \mathbf{x1},\text{ }\mathbf{y1} \right)\] are the focuses through which line passes
\[\therefore \mathbf{m}\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{75}{}^\circ \]
\[\mathbf{75}{}^\circ \text{ }=\text{ }\mathbf{45}{}^\circ \text{ }+\text{ }\mathbf{30}{}^\circ \]
Applying the eqn:
We realize that the point (x, y) lies on the line with incline m through the decent point (x1, y1), if and provided that, its directions fulfill the condition \[\mathbf{y}\text{ }\text{ }\mathbf{-y1}\text{ }=\text{ }\mathbf{m}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{-x1} \right)\]
Then, at that point, \[\mathbf{y}\text{ }\text{ }\mathbf{-2}\surd \mathbf{3}\text{ }=\text{ }\left( \mathbf{2}\text{ }+\text{ }\surd \mathbf{3} \right)\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{-2} \right)\]
\[\mathbf{y}\text{ }\text{ }\mathbf{-2}\surd \mathbf{3}\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{-4}\text{ }+\text{ }\surd \mathbf{3}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{-2}\text{ }\surd \mathbf{3}\]
\[\mathbf{y}\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{-4}\text{ }+\text{ }\surd \mathbf{3}\text{ }\mathbf{x}\]
\[\left( \mathbf{2}\text{ }+\text{ }\surd \mathbf{3} \right)\text{ }\mathbf{x}\text{ }\text{ }\mathbf{-y}\text{ }\text{ }\mathbf{-4}\text{ }=\text{ }\mathbf{0}\]
∴ The condition of the line is \[\left( \mathbf{2}\text{ }+\text{ }\surd \mathbf{3} \right)\text{ }\mathbf{x}\text{ }\text{ }\mathbf{-y}\text{ }\text{ }\mathbf{-4}\text{ }=\text{ }\mathbf{0}.\]