Question:
The equation of one of the straight lines which passes through the point $(1,3)$ and makes an angles
$\tan ^{-1}(\sqrt{2})$ with the straight line, $y+1=3 \sqrt{2} x$ is
Correct Option: 1
Solution:
$y=m x+c$
$3=m+c$
$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$
$=6 m+\sqrt{2}=m-3 \sqrt{2}$
$=\sin =-4 \sqrt{2} \rightarrow m=\frac{-4 \sqrt{2}}{5}$
$==6 m-\sqrt{2}=m-3 \sqrt{2}$
$=7 m-2 \sqrt{2} \rightarrow m=\frac{2 \sqrt{2}}{7}$
According to options take $\mathrm{m}=\frac{-4 \sqrt{2}}{5}$
So $\mathrm{y}=\frac{-4 \sqrt{2} \mathrm{x}}{5}+\frac{3+4 \sqrt{2}}{5}$
$4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0$