# Equation of a Line – Normal Form

**Normal:**A normal to a line is a line segment drawn from a point perpendicular to the given line. Let

**be the length of the normal drawn from the origin to a line, which subtends an angle ø with the positive direction of x-axis as follows. Then, we have Cos ø = p/m à m = p/Cos ø And Sin ø = p/n à n = p/Sin ø The equation of line in intercept form is, x/m + y/n =1 x/(p/Cos ø) + y/(p/Sin ø) = 1 x Cos ø/p + y Sin ø/p =1**

*p***This is called the normal form of equation of the given line making the angle ø with the positive direction of x-axis and whose perpendicular distance from the origin is**

*x Cos ø + y Sin ø = p.***.**

*p***Converting the general equation of a line into normal form:**The equation of a straight line in general form is, ax+by+c=0 From the above equations, we have Cos ø = p/m and Sin ø = p/n From Trigonometric identity, we have Cos

^{2 }ø + Sin

^{2}ø=1 p

^{2}/m

^{2}+ p

^{2}/n

^{2}=1 Solving this equation gives us, P = mn/√(m

^{2}+n

^{2}), which is the perpendicular distance from the origin to the line x/m + y/n =1 (i.e., nx + my – mn=0) Thus, we have the distance from origin to the line ax+by+c=0, = |c|/√(a

^{2}+b

^{2}) Thus, for converting the given line into normal form, divide the equation ax+by+c=0 by √(a

^{2}+b

^{2}). Thus,

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*[a/ √(a*^{2}+b^{2})] x + [b/√(a^{2}+b^{2})] + c/√(a^{2}+b^{2})=0