Ar=1/4(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)),
=1/4(0(1-3)+(-1)(3-(-1))+0(-1-1)),
=1/4(0+(-4)+0),
=1/4|-4|=|-4|/4=1 sq unit

Solution:
Given, vertices of a triangle are (0, -1), (2, 1) and (0, 3).
Let A (0, -1), B (2, 1) and C (0, 3) be the vertices of the triangle.
Midpoints of the sides of the triangle are (1, 0), (0, 1) and (1, 2), respectively.
Let D (1, 0), E (0, 1) and F (1, 2) be the midpoints of the sides AB, AC and BC, respectively.

To find the area of the triangle formed by joining the mid-points of the sides of the triangle, we need to find the length of the sides of the new triangle.

Length of DE = $\sqrt{(1-0)^2 + (0-1)^2}$ = $\sqrt{2}$
Length of EF = $\sqrt{(1-0)^2 + (2-1)^2}$ = $\sqrt{2}$
Length of DF = $\sqrt{(0-1)^2 + (3-0)^2}$ = $\sqrt{10}$

Let S be the semi-perimeter of the triangle DEF. Then,

S = $\frac{\sqrt{2} + \sqrt{2} + \sqrt{10}}{2}$ = $\frac{\sqrt{2} + \sqrt{10}}{2}$

Using Heron's formula, we can find the area of the triangle DEF as follows:

Area of triangle DEF = $\sqrt{S(S-\sqrt{2})(S-\sqrt{2})(S-\sqrt{10})}$

= $\sqrt{\frac{\sqrt{2} + \sqrt{10}}{2} \cdot \frac{\sqrt{2} + \sqrt{10} - \sqrt{2}}{2} \cdot \frac{\sqrt{2} + \sqrt{10} - \sqrt{2}}{2} \cdot \frac{\sqrt{2} + \sqrt{10} - \sqrt{10}}{2}}$

= $\sqrt{\frac{\sqrt{2} + \sqrt{10}}{2} \cdot \frac{\sqrt{2} + \sqrt{10} - \sqrt{2}}{2} \cdot \frac{\sqrt{2} + \sqrt{10} - \sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}}$

= $\sqrt{\frac{\sqrt{2} + \sqrt{10}}{16} \cdot \sqrt{2}}$

= $\sqrt{\frac{\sqrt{5} + 1}{8}}$

= $\frac{1}{2}$

Therefore, the area of the triangle formed by joining the mid-points of the sides of the triangle, whose vertices are (0, -1), (2, 1) and (0, 3) is 1.

Hence, the correct option is (d).