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 $$x+y+\sqrt{3}(y-x)=a$$.

Answer 1

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

Here,

Equation of the given line is,

$$x+y+\sqrt{3}(y-x)=a$$

$$\Rightarrow (\sqrt{3}+1)y=(\sqrt{3}-1)x+a$$

$$\Rightarrow y=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}x+\frac{a}{(\sqrt{3}+1)}$$

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

we get, 

$$m=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$$

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

So, the equations of the required lines are

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

$$\Rightarrow y=\frac{2-\sqrt{3}+2+\sqrt{3}}{1-(2-\sqrt{3})(2+\sqrt{3})}x\text{ and }y=\frac{2-\sqrt{3}-2-\sqrt{3}}{1+(2-\sqrt{3})(2+\sqrt{3})}x$$

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

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