Question

Derive the equation of straight line in the normal form x,cosθ + y, sinω = P, where Po is the length of normal to the line from the origin and ‘ω’ is inclination of normal with positive x-axis.

Answer 1

Derive the eqn. of straight line in the following

x cos ω + y sin ω = p (normal form)

consider a line cutting x-axis at A and y-axis at B.

Let the x-intercept = OA = a y-intercept OB = b

Drawn OM ⊥ to AB from 0

OM is called the length of the normal = P.

It makes an angle of the normal = P.

It makes an angle w (omega) w.r.t. X-axis.

In the angle AOM

cosω = OM/OA = P/a

∴ a = p/cosω

In the angle ROM

sinω = OM/OB = p/b

∴ b = P/sinω

Equation of the line in intercept is given by x/a + y/b = 1 substitute ‘a’ and ‘b’ then

x/(P/cosω) + y/(P/sinω) = 1

∴ [xcosα w + ysin α = p] in the equation of line AB is normal form.