If $(1, 2), (3, 4)$ and $(0, 6)$ are the three vertices of a parallelogram taken in that order, then the fourth vertex is .......... .

Let the vertices of the parallelogram be $A(1,2)$, $B(3,4)$, $C(0,6)$ and $D(x,y)$.

Thus, the diagonals are AC and BD.

Now diagonals of a parallelogram bisect each other.

Therefore, the midpoint of AC is the midpoint of BD.

Now the midpoint of AC is $\left(\dfrac{1+0}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{1}{2},4\right)$.

Now the midpoint of BD will be $\left(\dfrac{3+x}{2},\dfrac{4+y}{2}\right)=\left(\dfrac{1}{2},4\right)$

$\dfrac{3+x}{2}=\dfrac{1}{2}$ or $x=-2$.

Similarly $\dfrac{4+y}{2}=4$ or $y=4$

Hence the last vertex is $(-2,4)$.