The equation of one of the


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

  1. (1) $4 \sqrt{2} \mathrm{x}+5 \mathrm{y}-(15+4 \sqrt{2})=0$

  2. (2) $5 \sqrt{2} \mathrm{x}+4 \mathrm{y}-(15+4 \sqrt{2})=0$

  3. (3) $4 \sqrt{2} x+5 y-4 \sqrt{2}=0$

  4. (4) $4 \sqrt{2} \mathrm{x}-5 \mathrm{y}-(5+4 \sqrt{2})=0$

Correct Option: 1


$y=m x+c$


$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$

$=6 m+\sqrt{2}=m-3 \sqrt{2}$

$=\sin =-4 \sqrt{2} \rightarrow \mathrm{m}=\frac{-4 \sqrt{2}}{5}$

$=6 m-\sqrt{2}=m-3 \sqrt{2}$

$=7 \mathrm{~m}-2 \sqrt{2} \rightarrow \mathrm{m}=\frac{2 \sqrt{2}}{7}$

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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