Question

Find the perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

Answer 1

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given below:

$$\begin{gathered} y-\mathrm{sin\theta }=\frac{\mathrm{sin\varphi }-\mathrm{sin\theta }}{\mathrm{cos\varphi }-\mathrm{cos\theta }}\left(x-\mathrm{cos\theta }\right) \\ \Rightarrow \left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)y-\mathrm{sin\theta }\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)=\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)x-\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)\mathrm{cos\theta } \\ \Rightarrow \left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)x-\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)y+\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }=0 \end{gathered}$$

Let d be the perpendicular distance from the origin to the line $\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)\mathrm{x}-\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)\mathrm{y}+\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }=0$

$$\begin{gathered} \therefore d=\left|\frac{\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }}{\sqrt{{\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)}^2+{\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)}^2}}\right| \\ \Rightarrow d=\left|\frac{\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{{\sin }^2\mathrm{\varphi }+{\sin }^2\mathrm{\theta }-2\mathrm{sin\varphi sin\theta }+{\cos }^2\mathrm{\varphi }+{\cos }^2\mathrm{\theta }-2\cos \varphi \mathrm{cos\theta }}}\right| \\ \Rightarrow d=\left|\frac{\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{{\sin }^2\mathrm{\varphi }+{\cos }^2\mathrm{\varphi }+{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }-2\cos \left(\mathrm{\theta }-\mathrm{\varphi }\right)}}\right| \\ \Rightarrow d=\frac{1}{\sqrt{2}}\left|\frac{\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{1-\cos \left(\mathrm{\theta }-\mathrm{\varphi }\right)}}\right| \end{gathered}$$

$$\begin{gathered} \Rightarrow d=\frac{1}{\sqrt{2}}\left|\frac{\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{2{\sin }^2\left({\displaystyle \frac{\mathrm{\theta }-\mathrm{\varphi }}{2}}\right)}}\right| \\ \Rightarrow d=\frac{1}{\sqrt{2}\times \sqrt{2}}\left|\frac{\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sin \left({\displaystyle \frac{\mathrm{\theta }-\mathrm{\varphi }}{2}}\right)}\right|=\frac{1}{2}\left|\frac{2\sin \left({\displaystyle \frac{\mathrm{\theta }-\mathrm{\varphi }}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{\theta }-\mathrm{\varphi }}{2}}\right)}{\sin \left({\displaystyle \frac{\mathrm{\theta }-\mathrm{\varphi }}{2}}\right)}\right| \\ \Rightarrow d=\cos \left(\frac{\mathrm{\theta }-\mathrm{\varphi }}{2}\right) \end{gathered}$$

Hence, the required distance is $\cos \left(\frac{\mathrm{\theta }-\mathrm{\varphi }}{2}\right)$.