Find 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). Find the ratio of this area to the area of the given triangle.

The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :

D = \( (\frac{0+2}{2} \ , \ \frac{-1+1}{2}) \) = (1,0)
E = ( \( \frac{0+2}{2} \ , \ \frac{1+3}{2} \) ) = (1,2)
F = ( \( \frac{0+0}{2} \ , \ \frac{3-1}{2} \) ) = (0,1)

So, Area of \( ∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1 \)

∴ Area of ∆ DEF = 1 sq. units

Area of ∆ ABC = \( \frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4 \)

\(\therefore \) Area of \(\triangle \) ABC = 4 sq. units

So, \( ∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4 \).