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

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

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

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

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

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

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

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