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

Question

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

Smart Achievers · Accepted Answer

(1)

$Let the points be A = (\cos θ, \sin θ) and B = (\cos ϕ, \sin ϕ)$

$Equation of line A B is$

$y - \sin θ = \frac{\sin ϕ - \sin θ}{\cos ϕ - \cos θ} (x - \cos θ)$

$\Rightarrow$ $y - \sin θ = \frac{2 \cos (\frac{θ + ϕ}{2}) \sin (\frac{ϕ - θ}{2})}{- 2 \sin (\frac{ϕ + θ}{2}) \sin (\frac{ϕ - θ}{2})} (x - \cos θ)$

$\Rightarrow$ $y \sin (\frac{θ + ϕ}{2}) - \sin θ \sin (\frac{θ + ϕ}{2}) = - x \cos (\frac{θ + ϕ}{2}) + \cos θ \cos (\frac{θ + ϕ}{2})$

$\Rightarrow$ $x \cos (\frac{θ + ϕ}{2}) + y \sin (\frac{θ + ϕ}{2}) - {\cos θ \cos (\frac{θ + ϕ}{2}) + \sin θ \sin (\frac{θ + ϕ}{2})} = 0$

$\Rightarrow$ $x \cos (\frac{θ + ϕ}{2}) + y \sin (\frac{θ + ϕ}{2}) - \cos (θ - \frac{θ + ϕ}{2}) = 0$

$\Rightarrow$ $x \cos (\frac{θ + ϕ}{2}) + y \sin (\frac{θ + ϕ}{2}) - \cos (\frac{θ - ϕ}{2}) = 0$

$Now, perpendicular distance of line from the origin is$

$= \frac{| 0 + 0 - \cos (\frac{θ - ϕ}{2}) |}{\sqrt{\cos^{2} (\frac{θ + ϕ}{2}) + \sin^{2} (\frac{θ + ϕ}{2})}} =$ $| \cos (\frac{θ - ϕ}{2}) |$

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