Question

Find the equations to the straight lines passing through the point (3, −2) and inclined at 60° to the line $\sqrt{3}x+y=1$.

Answer 1

We know that the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-y_1=\frac{m\pm \tan \alpha }{1\mp m\tan \alpha }\left(x-x_1\right)$

Here,
$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}, \\ \sqrt{3}x+y=1 \\ \Rightarrow y=-\sqrt{3}x+1 \\ \mathrm{Comparing}\mathrm{this}\mathrm{equation}\mathrm{with}\mathit{}y=mx+c \\ \mathrm{we}\mathrm{get}, \\ m=-\sqrt{3} \end{gathered}$$
$x_1=3,y_1=-2,\mathrm{\alpha }={60}^{\circ },m=-\sqrt{3}$

So, the equations of the required lines are

$$\begin{gathered} y+2=\frac{{\displaystyle -\sqrt{3}+\tan {60}^{\circ }}}{{\displaystyle 1+\sqrt{3}\tan {60}^{\circ }}}\left(x-3\right)\mathrm{and}y+2=\frac{{\displaystyle -\sqrt{3}-\tan {60}^{\circ }}}{{\displaystyle 1-\sqrt{3}\tan {60}^{\circ }}}\left(x-3\right) \\ \Rightarrow y+2=\frac{{\displaystyle -\sqrt{3}+\sqrt{3}}}{{\displaystyle 1+3}}\left(x-3\right)\mathrm{and}y+2=\frac{{\displaystyle -\sqrt{3}-\sqrt{3}}}{{\displaystyle 1-3}}\left(x-3\right) \\ \Rightarrow y+2=0\mathrm{and}y+2=\sqrt{3}\left(x-3\right) \\ \Rightarrow y+2=0\mathrm{and}\sqrt{3}x-y=2+3\sqrt{3} \end{gathered}$$