Manipal Entrance Test 2014 Math Paper

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