Question

Angles made with the $X$ - axis by the two lines passing through the point $P(1,2)$ and cutting the line $x+y=4$ at a distance $\frac{\sqrt{6}}{3}$ units from the point $P$ are

• A. $\frac{\pi}{5}$ and $\frac{3 \pi}{10}$
• B. $\frac{\pi}{6}$ and $\frac{\pi}{3}$
• C. $\frac{\pi}{12}$ and $\frac{5 \pi}{12}$
• D. $\frac{\pi}{8}$ and $\frac{3 \pi}{8}$

Answer 1

Answer: (C)

Let slope of line is $m$
$\therefore $ Equation of line is
$y-2=m(x-1)$
$y-2=m x-m$
$m x-y+(2-m)=0$
On solving $m x-y+(2-m)$ and $x+y=4$, we get
$(x, y)=\left(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\right)$
Now, it is given that
$ \sqrt{\left(\frac{m+2}{m+1}-1\right)^{2}+\left(\frac{3 m+2}{m+1}-2\right)^{2}}=\frac{\sqrt{6}}{3}$
$\Rightarrow \frac{1}{(m+1)^{2}}+\frac{m^{2}}{(m+1)^{2}}=\frac{6}{9} $
$\Rightarrow \frac{1+m^{2}}{(m+1)^{2}}=\frac{2}{3}$
$\Rightarrow 3+3 m^{2}=2 m^{2}+4 m+2 $
$\Rightarrow m^{2}-4 m+1=0$
$ \Rightarrow m=2 \pm \sqrt{3} $
$\Rightarrow m=2+\sqrt{3}, 2-\sqrt{3}$
$\therefore $ Angles will be $\frac{\pi}{12}$ and $\frac{5 \pi}{12}$.