The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3) is:

1. (0, 1)

2. (0, -1)

3. (-1, 0)

4. (1, 0)

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

By mid-point formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values, we get :

For diagonal AC:

$\text{Mid-point of AC} = \Big(\dfrac{-2 + 8}{2}, \dfrac{3 + 3}{2}\Big) = \Big(\dfrac{6}{2}, \dfrac{6}{2}\Big) \\[1em] = (3, 3).$

Let point D be (x, y).

For diagonal BD:

$\text{Mid-point of BD} = \Big(\dfrac{6 + x}{2}, \dfrac{7 + y}{2}\Big) \\[1em] \Rightarrow (3, 3) = \Big(\dfrac{6 + x}{2}, \dfrac{7 + y}{2}\Big) \\[1em] \Rightarrow 3 = \dfrac{6 + x}{2} \text{ and } 3 = \dfrac{7 + y}{2} \\[1em] \Rightarrow 6 = 6 + x \text{ and } 6= 7 + y \\[1em] \Rightarrow x = 0 \text{ and } y = 6 - 7 = -1.$

D = (x, y) = (0, -1).

Hence, Option 2 is the correct option.