JEE Main 2021 (Online) 18th March Morning ShiftQuestion 1
The equation 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 :
Points and Straight Lines
Solution
Given slope $m_1=3\sqrt2$, required $\tan\phi=\sqrt2$: $\frac{m-3\sqrt2}{1+3\sqrt2m}=\pm\sqrt2$. Taking $m=-\frac{4\sqrt2}{5}$ (other root $\frac{2\sqrt2}{7}$): line through $(1,3)$: $y-3=-\frac{4\sqrt2}{5}(x-1)$. $\Rightarrow 4\sqrt2x+5y-(15+4\sqrt2)=0$. Option (A).