Question

Show that the equations to the straight lines passing through the point $(3,-2)$ and inclined at ${60}^o$ to the line $\sqrt{3}x+y=1$ are $y+2=0$ and $y-\sqrt{3}x+2+3\sqrt{3}=0$.

Answer 1

$\sqrt{3}x+y=1$

slope of line $=-\frac{a}{b}=-\frac{\sqrt{3}}{1}=-\sqrt{3}$

Angle between two line i.e. $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Let slope of line that passes through $(3,-2)$ be $m$

$\tan {60}^{\circ }=\left|\frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})}\right|\sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\pm \sqrt{3}\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\sqrt{3}m+\sqrt{3}=\sqrt{3}-3m\Rightarrow m=0\frac{m+\sqrt{3}}{1-\sqrt{3}m}=-\sqrt{3}m+\sqrt{3}=-\sqrt{3}+3m-2m=-2\sqrt{3}\Rightarrow m=\sqrt{3}$

Equtaion of line with slope $m=0$

$y+2=0(x-3)y+2=0$

Equation of line with slope $m=\sqrt{3}$

$y+2=\sqrt{3}(x-3)y-\sqrt{3}x+2+3\sqrt{3}=0$