Given: point (2, \( 2\sqrt{3}\)) and θ = 75°

Equation of line: (y – y_{1}) = m (x – x_{1})

where, m = slope of line = tan θ and (x_{1}, y_{1}) are the points through which line passes

m = tan 75°

75° = 45° + 30°

Applying the formula:

\( \dfrac{tan 45°+tan 30°}{1-tan 45°.tan 30°}\)

Substitute the values,

\(1+\dfrac{1}{\sqrt{3}}.1-\dfrac{1}{\sqrt{3}}\)

tan 45° = \(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

Let us rationalizing we get,

= \( \dfrac{3+1+2\sqrt{3}}{3-1}\)

= \( 2+\sqrt{3}\)

We know that the point (x, y) lies on the line with slope m through the fixed point (x_{1}, y1), if and only if, its coordinates satisfy the equation y – y_{1} = m (x – x_{1})

Then, y –\( 2\sqrt{3}\) = ( \( 2+\sqrt{3}\)) (x – 2)

y – \( 2\sqrt{3}\) = 2 x – 4 + \( \sqrt{3}x-\)\( 2\sqrt{3}\)

y = 2 x – 4 +\( \sqrt{3}x\)

( \( 2+\sqrt{3}\)) x – y – 4 = 0

The equation of the line is (\( 2+\sqrt{3}\)) x – y – 4 = 0.

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