A straight line $\mathrm{L}$ at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular drawn from the origin to this line makes an angle of $60^{\circ}$ with the line $x+y=0$. Then the equation of the line $\mathrm{L}$ is

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Aicharya · Accepted Answer

Given: $O P=4 \mathrm{~cm}$

Slope of $y+x=0$ is -1 . Slope of $O P$ is $\alpha$. Then, $60+\alpha=135^{\circ} \Rightarrow \alpha=75^{\circ}$ Now, $\cos 75^{\circ}=\cos \left(30^{\circ}+45^{\circ}\right)=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}$ $\sin 75^{\circ}=\sin \left(30^{\circ}+45^{\circ}\right)=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}$ Now, equation of line $\mathbf{L}$ is $\begin{aligned} & x \cos \alpha+y \sin \alpha=d \Rightarrow x \cos 75^{\circ}+y \sin 75^{\circ}=4 \\ & \Rightarrow x\left(\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)+y\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)=4 \\ & \Rightarrow(\sqrt{3}-1) x+(\sqrt{3}+1) y=8 \sqrt{2} \end{aligned}$

A straight line $\mathrm{L}$ at a distance of 4 units from the origin makes positive intercepts on the…

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