Question

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line $\sqrt{3}x+y=1$ are

(a) $y+2=0,\sqrt{3}x–y–2–3\sqrt{3}=0$

(b) $x–2=0,\sqrt{3}x–y+2+3\sqrt{3}=0$

(c) $\sqrt{3}x–y–2\pm 3\sqrt{3}=0$

(d) none of these

Answer 1

Given lines is $\sqrt{3}x+y=1$
Which has slope = $-\sqrt{3}$
Let us suppose slope of required line is m, which makes an angle of 60^∘ with above line.
$$\begin{gathered} \mathrm{i}.\mathrm{e}.\tan 60°=\left|\frac{-\sqrt{3}-m}{1-\sqrt{3}m}\right| \\ \mathrm{i}.\mathrm{e}.\sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right| \\ \mathrm{i}.\mathrm{e}.\pm \left(+\sqrt{3}-3m\right)=m+\sqrt{3} \\ \mathrm{i}.\mathrm{e}.\sqrt{3}-3m=m+\sqrt{3}\mathrm{or}-\sqrt{3}+3m=m+\sqrt{3} \\ \mathrm{i}.\mathrm{e}.m=0 \\ \mathrm{i}.\mathrm{e}.2m=2\sqrt{3} \\ \mathrm{i}.\mathrm{e}.m=\sqrt{3} \\ \therefore m=0\mathrm{or}m=\sqrt{3} \end{gathered}$$
Since line passes through (3, –2).
Equation of lien is y + 2 = $\sqrt{3}\left(x-3\right)$ or y + 2 = 0

$\mathrm{i}.\mathrm{e}.\sqrt{3}x-y-2-3\sqrt{3}=0\mathrm{or}y=-2$

Hence, the correct answer is option A.