If A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram ABCD, then find the values of x and y.

$x=6, y=3$

Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.
Mid point of diagonal AC is $((\frac{1+\mathrm{x)/}}{2},(\frac{2+6)/}{2})\Rightarrow ((\frac{x+\mathrm{1)/}}{2},4)$
Mid point of diagonal BD is$((\frac{4+\mathrm{3)/}}{2},(\frac{5+\mathrm{y)/}}{2})\Rightarrow (\frac{\mathrm{7/}}{2},(\frac{5+\mathrm{y)/}}{2})$
Since the diagonals of a parallelogram bisect each other, the midpoints of AC and BD are the same.
$\therefore (\frac{x+\mathrm{1)/}}{2}=\frac{\mathrm{7/}}{2}$ and $4=(\frac{5+\mathrm{y)/}}{2}$
$\Rightarrow x+1=7$ and $5+y=8$
$\Rightarrow x=6$ and $y=3$

The correct answer is Option (2) → $x=6, y=3$