The three vertices of a parallelogram are (3, 4) (3, 8)

We are given three vertices of a parallelogram A(3, 4),  B(3, 8) and C(9, 8). 

We know that diagonals of a parallelogram bisect each other. Let the fourth vertex be D(x, y).

Mid point of $B D=\left(\frac{x+3}{2}, \frac{y+8}{2}\right)$

Mid point of $A C=\left(\frac{9+3}{9}, \frac{4+8}{9}\right)=(6,6)$

Since the mid point of BD = mid point of AC

$\mathrm{So},\left(\frac{x+3}{2}, \frac{y+8}{2}\right)=(6,6)$

Thus, x = 9, y = 4.

So, the fourth vertex is (9, 4).