Question

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

Answer 1

Let the required equation be $y=mx+c$

But, c = 0 as it passes through origin (0, 0)

$\therefore$ Equation of the lines $y=mx+c$

Slope of $x+y+\sqrt{3}y=\sqrt{3}x=a$

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

$\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$

The angle between $x+y+\sqrt{3}y-\sqrt{3}=a$ and $y=mx$ is ${75}^{\circ }$.

$tan{45}^{\circ }=\frac{m_1\pm m_2}{1\mp m_1{m}_2}$

$tan({30}^{\circ }+{45}^{\circ })=\frac{m\pm (2-\sqrt{3})}{1-m(2-\sqrt{3})}$

$\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}\times 1}=\frac{m\pm (2-\sqrt{3})}{1-m(2-\sqrt{3})}$

$2+\sqrt{3}=\frac{m+2-\sqrt{3}}{1+m(\sqrt{3}-2)}$ and

$\therefore \frac{1}{m}=0$ or $m=-\sqrt{3}$

$\therefore y=mx$ $y=-\sqrt{3}x$ and $x=0$ are the required equations.