# Show that the equations to the straight lines passing through the point (3,−2) and inclined at 60@ to the line sqrt3x+y=1 are y+2=0 and y−sqrt3x+2+3sqrt3=0.

Show that the equations to the straight lines passing through the point $\left(3,-2\right)$ and inclined at ${60}^{\circ }$ to the line $\sqrt{3}x+y=1$ are $y+2=0$ and $y-\sqrt{3}x+2+3\sqrt{3}=0$
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vidovitogv5
Let $\theta$ be the angle between two lines whose slopes are ${m}_{1},{m}_{2}$. Then we have:
$\mathrm{tan}\theta =|\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}|$
$\mathrm{tan}\theta =|\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}|$
As we know ${m}_{1}$, we can just plug in the value and get ${m}_{2}$:
$\sqrt{3}=|\frac{-\sqrt{3}-{m}_{2}}{1-\sqrt{3}{m}_{2}}|$
$\sqrt{3}-3{m}_{2}=-\sqrt{3}-{m}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{m}_{2}=\sqrt{3}$
or
$-\sqrt{3}+3{m}_{2}=-\sqrt{3}-{m}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{m}_{2}=0$
$y+2=0\left(x-3\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y+2=0$
$y+2=\sqrt{3}\left(x-3\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y-\sqrt{3}x+2+3\sqrt{3}=0$