Question

Two lines passing through the point (2, 3) intersects each other at an angle of ${60}^{\circ }$. If slope of one line is 2; Find equation of the other line.

Answer 1

Here $m_1$ = 2 and $\theta ={60}^{\circ }$

We know that $tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\therefore tan{60}^{\circ }=\left|\frac{2-m_2}{1+2m_2}\right|$

$\Rightarrow \sqrt{3}=\left|\frac{2-m_2}{1+2m_2}\right|$

$\Rightarrow \frac{2-m_2}{1+2m_2}=\pm \sqrt{3}$

If $\frac{2-m_2}{1+2m_2}=\sqrt{3}$

$\Rightarrow 2-m_2=\sqrt{3}+2\sqrt{3}{m}_2$

$\Rightarrow (2\sqrt{3}+1)m_2=2-\sqrt{3}$

$\Rightarrow m_2=\frac{2-\sqrt{3}}{2\sqrt{3}+1}$

$\therefore$ Equation of required line is

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

$\Rightarrow (2\sqrt{3}+1)y-6\sqrt{3}-3$

$=(2-\sqrt{3})x-4+2\sqrt{3}$

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

$=-4+2\sqrt{3}+6\sqrt{3}+3$

$\Rightarrow (\sqrt{3}-2)x+(2\sqrt{3}+1)y=8\sqrt{3}-1$

If $\frac{2-m_2}{1+2m_2}=-\sqrt{3}$

$\Rightarrow 2-m_2=-\sqrt{3}-2\sqrt{3}{m}_2$

$\Rightarrow (2\sqrt{3}-1)m_2=-(2+\sqrt{3})$

$\Rightarrow m_2=\frac{-(2+\sqrt{3})}{(2\sqrt{3}-1)}$

$\therefore$ Equation of required line is

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

$\Rightarrow (2\sqrt{3}-1)y-6\sqrt{3}+3$

$=-(2+\sqrt{3})x+4+2\sqrt{3}$

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

$=4+2\sqrt{3}+6\sqrt{3}-3$

$\Rightarrow (2+\sqrt{3})x+\left(2\sqrt{3-1}\right)y=8\sqrt{3}+1.$