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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