Question

Find the equations of two straight lines passing through (1, 2) and making an angle of 60° with the line x + y = 0. Find also the area of the triangle formed by the three lines.

Answer 1

Let A(1, 2) be the vertex of the triangle ABC and x + y = 0 be the equation of BC.



Here, we have to find the equations of sides AB and AC, each of which makes an angle of ${60}^{\circ }$ with the line x + y = 0.

We know the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

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

Here,
$x_1=1,y_1=2,\mathrm{\alpha }={60}^{\circ },m=-1$

So, the equations of the required sides are

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


Solving x + y = 0 and $y-2=\left(2-\sqrt{3}\right)\left(x-1\right)$, we get:

$x=-\frac{\sqrt{3}+1}{2},y=\frac{\sqrt{3}+1}{2}$

$\therefore B\equiv \left(-\frac{\sqrt{3}+1}{2},\frac{\sqrt{3}+1}{2}\right)\mathrm{or}C\equiv \left(\frac{\sqrt{3}-1}{2},-\frac{\sqrt{3}-1}{2}\right)$



AB = BC = AD = $=\sqrt{6}\mathrm{units}$

$\therefore$ Area of the required triangle = $\frac{\sqrt{3}\times {\left(\sqrt{6}\right)}^2}{4}=\frac{3\sqrt{3}}{2}\mathrm{square}\mathrm{units}$