Question

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

Answer 1

Slope of the given line is $m_1=\frac{1-\sqrt{3}}{1+\sqrt{3}}$, $\theta ={75}^0$

$tan\theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$tan{75}^0=\left|\frac{m_2-\frac{1-\sqrt{3}}{1+\sqrt{3}}}{1+\frac{1-\sqrt{3}}{1+\sqrt{3}}{m}_2}\right|$..........(1)

As $tan{75}^0=2+\sqrt{3}$

From (1):

$\mp tan{75}^0=\frac{m_2-\frac{1-\sqrt{3}}{1+\sqrt{3}}}{1+\frac{1-\sqrt{3}}{1+\sqrt{3}}{m}_2}$

for $+tan{75}^0$ the value of $m_2=-\sqrt{3}$

And for $-tan{75}^0$ the value of $m_2=\infty$

Given the line passes through origin then equation of line will be $y=m_{2}x$

Therefore equations of the line are $y+\sqrt{3}x=0$ and $x=0$