Question

Find the perpendicular distance from the origin to the line joining the points $(cosθ,sinθ)$ and $(cos\phi,sin\phi).$

Answer 1

The equation of the line joining the points $(cosθ,sinθ)$ and $(cos\phi,sin\phi)$ is given by
$y-sinθ=\frac{sin\phi-sinθ}{cos\phi-cosθ}(x-cosθ)$

$y(cos\phi-cosθ)-sinθ(cos\phi-cosθ)=x(sin\phi-sinθ)-cosθ(sin\phi-sinθ)$

$x(sinθ-sin\phi)+y(cosθ-cos\phi)+sin(\phi-θ)=0$

$Ax+By+C=0$, where $A=sinθ-sin\phi,B=cos\phi-cosθ$ and $C=sin(\phi-θ)$

It is known that the perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}$
Therefore, the perpendicular distance (d) of the given line from point $(x_1, y_1)$ = (0, 0) is

$d=\frac{\left|(sinθ-sin\phi)(0)+(cos\phi-cosθ)(0)+sin(\phi-θ)\right|}{\sqrt{(sinθ-sin\phi)^2+(cos\phi-cosθ)^2}}$

$=\frac{|sin(\phi-θ)|}{\sqrt{sin^2θ+sin^2\phi-2sinθsin\phi+cos^2θ+cos^2\phi-2cosθcos\phi}}$

$=\frac{|sin(\phi-θ)|}{\sqrt{(sin^2θ+cos^2θ)+(sin^2\phi+cos^2\phi)-2(sinθsin\phi+cosθcos\phi)}}$

$=\frac{|sin(\phi-θ)|}{\sqrt{1+1-2(cos(\phi-θ))}}$

$=\frac{|sin(\phi-θ)|}{\sqrt{2(1-cos(\phi-θ)}}$

$=\frac{|sin(\phi-θ)|}{\sqrt{2(2sin^2(\frac{\phi-θ}{2}))}}$

$=\frac{|sin(\phi-θ)|}{|2sin(\frac{\phi-θ}{2})|}$