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.

Let A(0,-1), B(2,1), and C(0,3) be the vertices of ΔABC.
Area of ΔABC = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
Here (x1,y1) = (0,-1), (x2,y2) = (2,1), (x3,y3) = (0,3)
= 1/2[0(1-3) + 2(3+1) + 0(-1-1)]
= 8/2 = 4 sq. units
Let P, Q, R be the mid-points of AB, AC, and BC. Then coordinates of:
P = [(0+2)/2, (-1+1)/2] = [1, 0]
Q = [(0+0)/2, (-1+3)/2] = [0, 1]
R = [(2+0)/2, (1+3)/2] = [1, 2]
Area of ΔPQR = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
= 1/2[1(1-2) + 0(2-0) + 1(0-1)]
= -1/2 - 1/2 = -1 = 1 sq. unit (Area cannot be negative, take the absolute value)
Ratio of ΔABC and ΔPQR = 4:1