Let the equation of the line parallel to x−2y=1 is x−2y+λ=0 since, it passes through (3,5) ∴ 3−10+λ=0⇒λ=7 ∴ The line is x - 2y + 7 = 0, The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4) ∴ The distance between (3,5) and (1,4) =√(3−1)2+(5−4)2=√4+1=√5