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 , then the equation of other line is
I. $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y-8 \sqrt{3}-1=0$
II. $(2-\sqrt{3}) x-(1+2 \sqrt{3}) y+8 \sqrt{3}-1=0$

• A. Both I and II are possible
• B. Only I is possible
• C. Only II is possible
• D. Neither I nor II is possible

Answer 1

Answer: (A)

Equation of line $l_1$ by using $y-y_1=m\left(x-x_1\right)$
where, $\left(x_1, y_1\right)=(2,3)$ and $m=2$, is
$ y-3 =2(x-2) $
$\Rightarrow y-3 =2 x-4$
$\Rightarrow 2 x-y-1 =0 $
$\Rightarrow y =2 x-1 ....$(i)
Here, slope of line $l_1$ is $m_1=2$.
Let slope of line $l_2$ is $m$.

Given, $\theta-60^{\circ}$
Again taking negative sign, we get
$ \frac{2-m}{1+2 m}=-\sqrt{3} $
$\Rightarrow 2-m=-\sqrt{3}(1+2 m) $
$ \Rightarrow 2-m=-\sqrt{3}-2 \sqrt{3} m$
$ \Rightarrow m-2 \sqrt{3} m=\sqrt{3}+2 $
$\Rightarrow -m(2 \sqrt{3}-1)=(\sqrt{3}+2)$
$\Rightarrow m=-\left(\frac{\sqrt{3}+2}{2 \sqrt{3}-1}\right) $
Now, equation of line $l_2$ by using
$ y-y_1=m\left(x-x_1\right) \text { is }$
$y-3=\frac{2-\sqrt{3}}{1+2 \sqrt{3}}(x-2) $
$ \left(\text { when } m=\frac{2-\sqrt{3}}{1+2 \sqrt{3}} \text { and } x_1=2, y_1=3\right)$
$\Rightarrow y(1+2 \sqrt{3})-3-6 \sqrt{3}=x(2-\sqrt{3})-4+2 \sqrt{3} $
$\Rightarrow x(2-\sqrt{3})-y(1+2 \sqrt{3})+8 \sqrt{3}-1=0 $
Again, $ y-3=-\left(\frac{\sqrt{3}+2}{2 \sqrt{3}-1}\right)(x-2) $
$\Rightarrow (2 \sqrt{3}-1)-6 \sqrt{3}+3=-(\sqrt{3}+2) x+2 \sqrt{3}+4$
$\Rightarrow (2+\sqrt{3}) x+(2 \sqrt{3}-1) y-8 \sqrt{3}-1=0$