Mathematics - Straight Lines Question with Solution | TestHub

$y=mx+c$

$3=m+c$

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

$\Rightarrow 6m+\sqrt{2}=\pm \left(m-3\sqrt{2}\right)$

from Positive sign

$\Rightarrow 5m=-4\sqrt{2} \Rightarrow m=\frac{-4\sqrt{2}}{5}$

from negative sign

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

$\Rightarrow 7 \mathrm{m}=2\sqrt{2}\Rightarrow \mathrm{m}=\frac{2\sqrt{2}}{7}$

According to options take $m=\frac{-4\sqrt{2}}{5}$

So $y=\frac{-4\sqrt{2}x}{5}+\frac{3+4\sqrt{2}}{5}$

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