Calculus
posted by eddie .
Write the equation 4x + 5y – 3 = 0 in normal form. Then, find the length of the normal and the angle it makes with the positive x–axis.

From
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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