Question

Find perpendicular distance of the line joining the points $(cos \,\theta, sin \,\theta)$ and $(cos\, \phi, sin \,\phi)$ from the origin.

• A. $\left|cos\left(\frac{\theta-\phi}{2}\right)\right|$
• B. $\left|cos\left(\frac{\theta+\phi}{2}\right)\right|$
• C. $\left|sin\left(\frac{\theta+\phi}{2}\right)\right|$
• D. None of these

Answer 1

Answer: (A)

$\left|cos\left(\frac{\theta-\phi}{2}\right)\right|$

Answer 2

Let the points be $A = \left(cos\, \theta, sin\, \theta\right)$ and $B = \left(cos\,\phi, sin\,\phi\right)$. Equation of line $AB$ is $y-sin\,\theta=\frac{sin\, \phi-sin\,\theta}{cos\,\phi-cos\,\theta}\left(x-cos\,\phi\right)$ or $y-sin\,\theta=\frac{2\,cos\left(\frac{\theta+\phi}{2}\right)sin\left(\frac{\phi-\theta}{2}\right)}{-2\, sin\left(\frac{\phi+\theta}{2}\right)sin\left(\frac{\phi-\theta}{2}\right)}\left(x-cos\, \theta\right)$ or $y\, sin\left(\frac{\theta+\phi}{2}\right)-sin\, \theta\,sin\left(\frac{\theta+\phi}{2}\right)$ $=-x\,cos\left(\frac{\theta+\phi}{2}\right)+cos\,\theta\,cos\left(\frac{\theta+\phi}{2}\right)$ or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)$ -\left\\{cos\,\theta\,cos\left(\frac{\theta +\phi }{2}\right)+sin\,\theta\,sin\left(\frac{\theta +\phi }{2}\right)\right\\}=0 or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)-cos\left(\theta-\frac{\theta +\phi }{2}\right)=0$ or $x\,cos\left(\frac{\theta +\phi }{2}\right)+y\,sin\left(\frac{\theta +\phi }{2}\right)-cos\left(\frac{\theta -\phi }{2}\right)=0$ $\therefore$ Perpendicular distance of line from the origin $=\frac{\left|0+0-cos\left(\frac{\theta-\phi}{2}\right)\right|}{\sqrt{cos^{2}\left(\frac{\theta+\phi}{2}\right)+sin^{2}\left(\frac{\theta+\phi}{2}\right)}}$ $=\left|cos\left(\frac{\theta-\phi}{2}\right)\right|$