JEE Main 2019 (Online) 12th April Evening ShiftQuestion 1
A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^{\circ}$ with the line $x+y=0$. Then an equation of the line $L$ is :
Points and Straight Lines
Solution
Write $L$ in normal form $x\cos\alpha+y\sin\alpha=4$, where $\alpha$ is the angle of the perpendicular $OM$ with the $x$-axis. Positive intercepts give $0<\alpha<\frac\pi2$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 225"><path stroke-width=".8" d="M12 147.7h186M67.3 213.1V12"/><path fill="#000" stroke="none" d="m198 147.7-5-2.2v4.4zM67.3 12l-2.2 5h4.4z"/><text x="203.6" y="157.5" fill="#000" stroke="none" font-style="italic">x</text><text x="73.3" y="15.5" fill="#000" stroke="none" font-style="italic">y</text><text x="61.7" y="157.5" fill="#000" stroke="none" font-style="italic" text-anchor="end">O</text><path stroke-dasharray="4 3" d="m12 92.4 120.6 120.7"/><text x="33.5" y="110.4" fill="#000" stroke="none" font-size="8.5">x+y=0</text><path stroke-width="1.2" d="M188.9 213.1 135 12"/><text x="151.7" y="51.7" fill="#000" stroke="none" font-size="8.5">L</text><path d="m67.3 147.7 97.1-26"/><circle cx="164.4" cy="121.7" r="1.9" fill="#000" stroke-width=".7"/><text x="170" y="131.5" fill="#000" stroke="none">M</text><path stroke-width=".8" d="m158.6 123.3-1.6-5.8 5.8-1.6M88.5 142a22 22 0 0 1-5.6 21.3"/><text x="97.2" y="159.3" fill="#000" stroke="none" font-size="8.5" text-anchor="middle">60°</text><text x="115.1" y="130.4" fill="#000" stroke="none" font-size="8.5" text-anchor="middle">4</text></svg> The line $x+y=0$ points along $-45^\circ$. $OM$ makes $60^\circ$ with it, so $\alpha=-45^\circ+60^\circ=15^\circ$ (the other choice, $\alpha=75^\circ$, gives $(\sqrt3-1)x+(\sqrt3+1)y=8\sqrt2$, which is not offered). $\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}$, $\sin15^\circ=\frac{\sqrt3-1}{2\sqrt2}$, so $\frac{\sqrt3+1}{2\sqrt2}x+\frac{\sqrt3-1}{2\sqrt2}y=4\Rightarrow\left(\sqrt3+1\right)x+\left(\sqrt3-1\right)y=8\sqrt2$. Answer (A).