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July 30, 2014

July 30, 2014

Posted by **eddie** on Tuesday, May 8, 2012 at 11:42am.

- Calculus -
**MathMate**, Tuesday, May 8, 2012 at 1:30pmFrom

4x + 5y – 3 = 0

Divide all coefficients by sqrt(4^2+5^2) to get

4x/sqrt(41) +5y/sqrt(41) - 3/sqrt(41) = 0

where

3/sqrt(41) = length of normal

cos(θ)=4/sqrt(41), and

sin(θ)=5/sqrt(41)

Note: the normal form is given by:

x cos(θ) + y sin(θ) - p =0

where |p|=length of normal.

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