# 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

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