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
$y-3=2(x-2)$
$\Rightarrow y-3=2x-4$
$\Rightarrow 2x-y-1=0$
$\Rightarrow y=2x-\mathrm{1....}$(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
$\frac{2-m}{1+2m}=-\sqrt{3}$
$\Rightarrow 2-m=-\sqrt{3}(1+2m)$
$\Rightarrow 2-m=-\sqrt{3}-2\sqrt{3}m$
$\Rightarrow m-2\sqrt{3}m=\sqrt{3}+2$
$\Rightarrow -m(2\sqrt{3}-1)=(\sqrt{3}+2)$
$\Rightarrow m=-\left(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\right)$
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, $y-3=-\left(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\right)(x-2)$
$\Rightarrow (2\sqrt{3}-1)-6\sqrt{3}+3=-(\sqrt{3}+2)x+2\sqrt{3}+4$
$\Rightarrow (2+\sqrt{3})x+(2\sqrt{3}-1)y-8\sqrt{3}-1=0$