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If the perpendicular distance of the straight line {\frac  {x}{a}}+{\frac  {y}{b}}=1\,is p, prove that {\frac  {1}{p^{2}}}={\frac  {1}{a^{2}}}+{\frac  {1}{b^{2}}}\,.

Wrting the given equation in general form,i tlooks like


The perpendicular distance from the origin to this equation is

|{\frac  {b\cdot 0+a\cdot 0-ab}{{\sqrt  {a^{2}+b^{2}}}}}|=p\,

{\frac  {ab}{{\sqrt  {a^{2}+b^{2}}}}}=p\,

Taking inverse on both sides and squaring,we get

{\frac  {a^{2}+b^{2}}{a^{2}+b^{2}}}={\frac  {1}{p^{2}}}\,

{\frac  {1}{b^{2}}}+{\frac  {1}{a^{2}}}={\frac  {1}{p^{2}}}\,

Hence proved

Main Page:Geometry:Straight Lines-I