Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x – 2y = 1.

Given: (x₁,y₁) = A(3, 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 x – 2y = 1

⇒ 2y = x – 1

⇒ y

⇒ sin θ and cos θ

So, the equation of the line is

Formula Used:

⇒

⇒ x – 2y + 7 = 0

Let x – 2y + 7 = 0 intersect the line 2x + 3y = 14 at point P.

Let AP = r

Then, the coordinate of P is given by

⇒ x and y

Thus, the coordinate of P is

Clearly, P lies on the line 2x + 3y = 14

⇒

⇒

⇒ r

Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is