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Find the equation of the line passing thro'the point (2,{\frac  {\pi }{6}})\, parallel and perpendicular to the line {\frac  {3}{r}}=4\cos \theta -3\sin \theta \,

Given line is {\frac  {3}{r}}=4\cos \theta -3\sin \theta \,

Any line parallel to this line is {\frac  {k}{r}}=4\cos \theta -3\sin \theta \,

This line passes thro' (2,{\frac  {\pi }{6}})\,

Therefore, {\frac  {k}{2}}=4\cos {\frac  {\pi }{6}}-3\sin {\frac  {\pi }{6}}\,

k=4{\sqrt  {3}}-3\,

Hence the line is {\frac  {4{\sqrt  {3}}-3}{r}}=4\cos \theta -3\sin \theta \,

Any line perpendicular to the given line is

{\frac  {k_{1}}{r}}=-4\sin \theta -3\cos \theta \,

Given this line passes thro' (2,{\frac  {\pi }{6}})\,

{\frac  {k_{1}}{2}}=-4\sin {\frac  {\pi }{6}}-3\cos {\frac  {\pi }{6}}\,

k_{1}=-(4+3{\sqrt  {3}})\,

The required line is {\frac  {(4+3{\sqrt  {3}})}{r}}=4\sin \theta +3\cos \theta \,

Main Page:Geometry:Polar Coordinates