Geo2.25

Change the equation $ax+by+c=0,(a^2+b^2)\ne 0\,$ 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) $c\ne 0\,$

$ax+by=-c\,$

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

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

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

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