Question

Find the equation of the line straight which pass through the origin and are inclined at an angle of ${75}^0$ to the straight line $x+y+\sqrt{3}(y-x)=a$

Answer 1

$x+y+\sqrt{3}(y-x)=ax-\sqrt{3}x+\sqrt{3}y+y=ax(1-\sqrt{3})+y(1+\sqrt{3})=ay(1+\sqrt{3})=a-x(1-\sqrt{3})y=-x\frac{(1-\sqrt{3})}{(1+\sqrt{3})}+\frac{a}{(1+\sqrt{3})}\tan 75=\left|\frac{\frac{1}{2+\sqrt{3}}-m_2}{1+\frac{1}{2+\sqrt{3}}.m_2}\right|\tan 75=\frac{1-({2m}_2+\sqrt{3}{m}_2)}{2+{\sqrt{3}+m}_2}$
$\tan 75=\frac{1-{2m}_2+\sqrt{3}{m}_2}{2+{\sqrt{3}+m}_2}-\tan 75=\frac{1-{2m}_2+\sqrt{3}{m}_2}{2+{\sqrt{3}+m}_2}-{\tan }^{2}75-\tan 75\times m_2=1-\tan 75\times m_2$
${\tan }^{2}75+m_2\tan 75=1-m_2\tan 75{\tan }^{2}75+2m_2\tan 75=1\tan 75(\tan 75+2m_2)=12m_2=\frac{1}{\tan 75}-\tan 75m_2=-1.732m_2=-\sqrt{3}$
Hence equation of line.
$y-0=-\sqrt{3}(x-0)y=-\sqrt{3}x$