Question

Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line \[\sqrt{3}x + y = 11\].

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, 

\[x_1 = 0, y_1 = 0, \alpha = {45}^\circ \text { and } m = - \sqrt{3}\]

So, the equations of the required lines are

\[y - 0 = \frac{- \sqrt{3} + \tan {45}^\circ}{1 + \sqrt{3}\tan {45}^\circ}\left( x - 0 \right) \text { and } y - 0 = \frac{- \sqrt{3} - \tan {45}^\circ}{1 - \sqrt{3}\tan {45}^\circ}\left( x - 0 \right)\]

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

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

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