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 \[\alpha\] 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, 

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

So, the equations of the required sides are

\[y - 2 = \frac{- 1 + \tan {60}^\circ}{1 + \tan {60}^\circ}\left( x - 1 \right) \text { and } y - 2 = \frac{- 1 - \tan {60}^\circ}{1 - \tan {60}^\circ}\left( x - 1 \right)\]

\[ \Rightarrow y - 2 = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\left( x - 1 \right)\text {  and } y - 2 = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\left( x - 1 \right)\]

\[ \Rightarrow y - 2 = \left( 2 - \sqrt{3} \right)\left( x - 1 \right) \text { and } y - 2 = \left( 2 + \sqrt{3} \right)\left( x - 1 \right)\]

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

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

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

AB = BC = AD = \[= \sqrt{6} \text { units }\]

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