Geo5.4.7

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Prove that the points ({\frac  {3}{2}},0),(-{\sqrt  {2}},45^{\circ }),(-{\frac  {3}{5}},90^{\circ })\, are collinear.

Area of the triangle formed is

{\frac  {1}{2}}[{\frac  {3}{2}}(-{\sqrt  {2}})\sin(-45)+(-{\sqrt  {2}}(-{\frac  {3}{5}})\sin(45-90)-{\frac  {3}{5}}{\frac  {3}{2}}\sin 90]\,

{\frac  {1}{2}}[{\frac  {3}{2}}({\sqrt  {2}})(-{\frac  {1}{{\sqrt  {2}}}})+{\frac  {3{\sqrt  {2}}}{5}}{\frac  {-1}{{\sqrt  {2}}}}-{\frac  {9}{10}}]\,

{\frac  {1}{2}}[{\frac  {15-6-9}{10}}]=0\,

Therefore,the points are collinear.


Main Page:Geometry:Polar Coordinates