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First find the line parallel to x-2y = 1 passing through (3,5) It will be x-2y = c put(3,5) in the line equation as it is a point on the line. 3- 2×5 = c ⇒c = -7 Equation of line : x - 2y = -7

Now find point of intersection of (x-2y=-7) and (2x+3y-14=0). You can find it by solving the 2 equations of line.

x - 2y + 7 = 0 <<<<<(1) 2x + 3y -14 = 0 <<<<<(2) 2x - 4y +14 = 0 <<<<<(3)(by multiplying 2 with eqn (1)) By Subtracting (3) from (2) 7y -28 =0 ⇒ y = 4 Put y=4 in eqn(1) x = 1

So the point of intersection is (1.4). Find the distance between (3,5) and (1.4).That is the required answer. Distance = √(3-1)²+(5-4)² = √2²+1² = √5