Given:

p = 3, α = tan^{-1} (\( \dfrac{5}{12}\))

So, tan α = \( \dfrac{5}{12}\)

sin α = \( \dfrac{5}{13}\)

cos α = \( \dfrac{12}{13}\)

The equation of the line in normal form is given by

By using the formula,

x cos α + y sin α = p

Now, substitute the values, we get

\( \dfrac{12x}{13}+ \dfrac{5y}{13}=3\)

12x + 5y = 39

The equation of line in normal form is 12x + 5y = 39.

Answered by Aaryan | 1 year agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0

Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0