# Geo2.25

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

$[{\frac {-ap}{c}}]^{{2}}+[{\frac {-bp}{c}}]^{2}=1\,hence[itex]p={\frac {pmc}{{\sqrt {a^{2}+b^{2}}}}}\,$

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

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