Question

Find the equations of two straight lines passing through (1, 2) and making an angle of ${60}^{\circ }$ 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 that equations of two lines passing through a point $(x_1,y_1)$ and making an angle $\alpha$ with the line whose slope is m.

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

Here,

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

So, the equations of the required sides are

$y-2=\frac{-1+tan{60}^{\circ }}{1+tan{60}^{\circ }}(x-1)$ and $y-2$

$=\frac{1-tan{60}^{\circ }}{1-tan{60}^{\circ }}(x-1)$

$\Rightarrow y-2=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x-1)$ and $y-2=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x-1)$

$\Rightarrow y-2=(2-\sqrt{3})(x-1)$ and $y-2=(2+\sqrt{3})(x-1)$

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

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

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

$C\equiv (\frac{\sqrt{3}-1}{2},-\frac{\sqrt{3}+1}{2})$

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

Area of the required triangle

$\therefore =\frac{\sqrt{3}\times (\sqrt{6}{)}^2}{4}=\frac{3\sqrt{3}}{2}$ squre units.