Geo2.25

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Change the equation ax+by+c=0,(a^{2}+b^{2})\neq 0\, into normal form

Case(i) c=0

In this case,the straight line passes through the origin and hence,p=0.

Now dividing the equation ax+by+c=0 by

{\sqrt  {a^{2}+b^{2}}}\,

we get

{\frac  {a}{{\sqrt  {a^{2}+b^{2}}}}}x+{\frac  {b}{{\sqrt  {a^{2}+b^{2}}}}}y=0\,

This can be interpreted as

\cos \alpha ={\frac  {a}{{\sqrt  {a^{2}+b^{2}}}}},\sin \alpha ={\frac  {b}{{\sqrt  {a^{2}+b^{2}}}}}\,

Hence

x\cos \alpha +y\sin \alpha =0\,

Case(ii) c\neq 0\,

ax+by=-c\,

Let the normal form be

x\cos \alpha +y\sin \alpha =p\,

Since the above two equations represent the same line,their slopes are equal.

Now we have,

{\frac  {\cos \alpha }{a}}={\frac  {\sin \alpha }{b}}={\frac  {p}{-c}}\,

From

\cos ^{2}\alpha +\sin ^{2}\alpha =1\,

Now

[{\frac  {-ap}{c}}]^{{2}}+[{\frac  {-bp}{c}}]^{2}=1\,</,math>hence<math>p={\frac  {pmc}{{\sqrt  {a^{2}+b^{2}}}}}\,

Now

\cos \alpha ={\frac  {a}{{\sqrt  {a^{2}+b^{2}}}}},\sin \alpha ={\frac  {b}{{\sqrt  {a^{2}+b^{2}}}}}\,

Writing these values in normal form equation

we get the normal form as

{\frac  {a}{{\sqrt  {a^{2}+b^{2}}}}}+{\frac  {b}{{\sqrt  {a^{2}+b^{2}}}}}={\frac  {-c}{{\sqrt  {a^{2}+b^{2}}}}}\,


Main page:Geometry:Straight Lines-I