<s>[INST] A coordinate geometry question. On pg. 46 of "Coordinate geometry" by S.L. Loney, the following question has been posed:

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

I am getting different equations for those two lines. 
My method:
$L_1: \frac{y+2}{x-3}=\frac{-\sqrt{3}+\tan(60^\circ)}{1-(-\sqrt{3})\tan(60^\circ)}$
$L_2: \frac{y+2}{x-3}=\frac{-\sqrt{3}-\tan(60^\circ)}{1+(-\sqrt{3})\tan(60^\circ)}$
On simplifying, $L_1$ and $L_2$ are coming out different as compared to the given equations. 
Any help would be much appreciated. 
 [/INST] Let $\theta$ be the angle between two lines whose slopes are $m_1,m_2$. Then we have:
$$\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}|$$
As we know $m_1$, we can just plug in the value and get $m_2$:
$$\sqrt{3} = |\frac{-\sqrt{3}-m_2}{1-\sqrt{3}{m_2}}|$$
$$\sqrt{3}-3m_2 = -\sqrt{3}-m_2 \implies m_2 = \sqrt{3}$$ or 
$$-\sqrt{3}+3m_2 = -\sqrt{3}-m_2 \implies m_2 = 0$$
$L_2$: $y+2 = 0(x-3) \implies y+2=0$
$L_1$: $y+2 = \sqrt{3}(x-3) \implies y-\sqrt{3}x+2+3\sqrt{3} = 0$
</s>