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.

(1:4) Let the vertices of the triangle be $A(0,-1), B(2,1), C(0,3)$.
Let $D, E, F$ be the mid-points of the sides $AB, AC, BC$ respectively. The coordinates of $D, E,$ and $F$ are calculated using the mid-point formula:
$D = \left(\frac{0+2}{2}, \frac{-1+1}{2}\right) = (1,0)$
$E = \left(\frac{0+0}{2}, \frac{-1+3}{2}\right) = (0,1)$
$F = \left(\frac{2+0}{2}, \frac{1+3}{2}\right) = (1,2)$
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area of $\Delta DEF = \frac{1}{2} |1(1-2) + 0(2-0) + 1(0-1)| = \frac{1}{2} |-1 + 0 - 1| = \frac{1}{2} |-2| = 1$ square unit.
Area of $\Delta ABC = \frac{1}{2} |0(1-3) + 2(3-(-1)) + 0(-1-1)| = \frac{1}{2} |0 + 2(4) + 0| = \frac{1}{2} |8| = 4$ square units.
Therefore,the required ratio is $1:4$.