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
$y=m x+c$
$3=m+c$
$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$
$=6 m+\sqrt{2}=m-3 \sqrt{2}$
$=\sin =-4 \sqrt{2} \rightarrow \mathrm{m}=\frac{-4 \sqrt{2}}{5}$
$=6 m-\sqrt{2}=m-3 \sqrt{2}$
$=7 \mathrm{~m}-2 \sqrt{2} \rightarrow \mathrm{m}=\frac{2 \sqrt{2}}{7}$
According to options take $\mathrm{m}=\frac{-4 \sqrt{2}}{5}$
So $y=\frac{-4 \sqrt{2} x}{5}+\frac{3+4 \sqrt{2}}{5}$
$4 \sqrt{2} \mathrm{x}+5 \mathrm{y}-(15+4 \sqrt{2})=0$
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