Question

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is 15°.

Answer 1

Here, p = 4, $\mathrm{\alpha }={15}^{\circ }$

$$\begin{gathered} \mathrm{Now},\cos {15}^{\circ }=\cos \left({45}^{\circ }-{30}^{\circ }\right)=\cos {45}^{\circ }\cos {30}^{\circ }+\sin {45}^{\circ }\sin {30}^{\circ } \\ \Rightarrow \cos {15}^{\circ }=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}} \end{gathered}$$

$$\begin{gathered} \mathrm{And},\sin {15}^{\circ }=\sin \left({45}^{\circ }-{30}^{\circ }\right)=\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ } \\ \Rightarrow \sin {15}^{\circ }=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}} \end{gathered}$$

So, the equation of the line in normal form is

$$\begin{gathered} x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p \\ \Rightarrow \frac{\left(\sqrt{3}+1\right)x}{2\sqrt{2}}+\frac{\left(\sqrt{3}-1\right)y}{2\sqrt{2}}=4 \\ \Rightarrow \left(\sqrt{3}+1\right)x+\left(\sqrt{3}-1\right)y=8\sqrt{2} \end{gathered}$$