Question

The equations of one of the straight lines which passes through the point $(1,3)$ and makes an angle ${\tan }^{-1}\sqrt{2}$ with the straight line, $y+1=3\sqrt{2}x$ is:

• A. $5\sqrt{2}x+4y-(15+4\sqrt{2})=0$
• B. $4\sqrt{2}x-5y-(5+4\sqrt{2})=0$
• C. $4\sqrt{2}x+5y-4\sqrt{2}=0$
• D. $4\sqrt{2}x+5y-(15+4\sqrt{2})=0$

Answer 1

The correct option is D $4\sqrt{2}x+5y-(15+4\sqrt{2})=0$

$\tan \left({\tan }^{-1}\sqrt{2}\right)=\left|\frac{m-3\sqrt{2}}{1+3m\sqrt{2}}\right|$
$\Rightarrow \sqrt{2}=\left|\frac{m-3\sqrt{2}}{1+3m\sqrt{2}}\right|$

Case $1:\left|\frac{m-3\sqrt{2}}{1+3m\sqrt{2}}\right|$ is $+ve$
$\Rightarrow 6m+\sqrt{2}=m-3\sqrt{2}$
$\Rightarrow 5m=-4\sqrt{2}$
$\Rightarrow m=-\frac{4\sqrt{2}}{5}$

Case $2:\left|\frac{m-3\sqrt{2}}{1+3m\sqrt{2}}\right|$ is $-ve$

$\Rightarrow -6m-\sqrt{2}=m-3\sqrt{2}$
$\Rightarrow 2\sqrt{2}=7m$
$\Rightarrow m=\frac{2\sqrt{2}}{7}$