Question

Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75° to the straight line \[x + y + \sqrt{3}\left( y - x \right) = a\].

Answer 1

We know that the equations of two lines passing through a point \[\left( x_1 , y_1 \right)\] and making an angle \[\alpha\] with the given line y = mx + c are \[y - y_1 = \frac{m \pm \tan\alpha}{1 \mp m\tan\alpha}\left( x - x_1 \right)\]

Here,

Equation of the given line is,

\[x + y + \sqrt{3}\left( y - x \right) = a\]

\[ \Rightarrow \left( \sqrt{3} + 1 \right)y = \left( \sqrt{3} - 1 \right)x + a\]

\[ \Rightarrow y = \frac{\left( \sqrt{3} - 1 \right)}{\left( \sqrt{3} + 1 \right)}x + \frac{a}{\left( \sqrt{3} + 1 \right)}\]

\[\text { Comparing this equation with } y = mx + c\]

we get, 

\[m = \frac{\left( \sqrt{3} - 1 \right)}{\left( \sqrt{3} + 1 \right)}\]

\[\therefore x_1 = 0, y_1 = 0, \alpha = {75}^\circ , m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}\] and \[\tan {75}^\circ = 2 + \sqrt{3}\]

So, the equations of the required lines are

\[y - 0 = \frac{2 - \sqrt{3} + \tan {75}^\circ}{1 - \left( 2 - \sqrt{3} \right)\tan {75}^\circ}\left( x - 0 \right) \text { and  }y - 0 = \frac{2 - \sqrt{3} - \tan {75}^\circ}{1 + \left( 2 - \sqrt{3} \right)\tan {75}^\circ}\left( x - 0 \right)\]

\[ \Rightarrow y = \frac{2 - \sqrt{3} + 2 + \sqrt{3}}{1 - \left( 2 - \sqrt{3} \right)\left( 2 + \sqrt{3} \right)}x \text { and } y = \frac{2 - \sqrt{3} - 2 - \sqrt{3}}{1 + \left( 2 - \sqrt{3} \right)\left( 2 + \sqrt{3} \right)}x\]

\[ \Rightarrow y = \frac{4}{1 - 1}x \text { and }y = - \sqrt{3}x\]

\[ \Rightarrow x = 0 \text { and }\sqrt{3}x + y = 0\]