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,

\[\alpha = {15}^\circ\]

\[\text { Now ,} \cos {15}^\circ = \cos\left( {45}^\circ - {30}^\circ \right) = \cos {45}^\circ \cos {30}^\circ + \sin {45}^\circ \sin {30}^\circ \]

\[ \Rightarrow \cos {15}^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}\]

\[\text {And,} \sin {15}^\circ = \sin\left( {45}^\circ - {30}^\circ \right) = \sin {45}^\circ \cos {30}^\circ - \cos {45}^\circ \sin {30}^\circ \]

\[ \Rightarrow \sin {15}^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}\]

So, the equation of the line in normal form is

\[xcos\alpha + ysin\alpha = p\]

\[ \Rightarrow \frac{\left( \sqrt{3} + 1 \right)x}{2\sqrt{2}} + \frac{\left( \sqrt{3} - 1 \right)y}{2\sqrt{2}} = 4\]

\[ \Rightarrow \left( \sqrt{3} + 1 \right)x + \left( \sqrt{3} - 1 \right)y = 8\sqrt{2}\]