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