Given: (x₁,y₁) = A(2,5)

To find:

The distance of a point from the line parallel to another line.

Explanation:

It is given that the required line is parallel to 3x −4y + 8 = 0

⇒ 4y = 3x + 8

⇒ y

∴ tanθ

⇒ sinθ, cosθ

So, the equation of the line is

⇒ 3x – 6 = 4y – 20

⇒ 3x – 4y + 14 = 0

Let the line 3x – 4y + 14 = 0 cut the line 3x + y + 4 = 0 at P.

Let AP = r Then, the coordinates of P are given by

⇒ x, y

Thus, the coordinates of P are

Clearly, P lies on the line 3x + y + 4 = 0.

∴

⇒

⇒

⇒ r = – 5

∴ AP = |r| = 5