Find the area of ΔABC with vertices A(0, -1), B(2,1) and C(0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4:1.

#### Solution

Let `A (x_1=0,y_1=-1), B (x_2 =2,y_2=1) and C (x_3 = 0, y_3=3) `be the given points. Then

`Area (Δ ABC)=1/2[x_1 (y_2-y_3)+ x_2(y_3-y_1) +x_3(y_1-y_3)]`

`=1/2 [0(1-3)+2(3+1)+0(-1-1)]`

`=1/2xx8=4` sq . units

So, the area of the triangle 4 ΔABC is sq units.

Let `D(a_1,b_1),E)a_2,b_2) and F (a_3,b_3)` be the midpoints of AB, BC and AC respectively

Then

`a_1 = (0+2)/2=1 b_2=(-1+1)/2=0`

`a_2=(2+0)/2=1 b_2=(1+3)/2=2`

`a_3=(0+0)/2=0 b_3 = (-1+3)/2=1`

Thus, the coordinates of D,E and F are D`(a_1=1,b_1=0),E(a_2=1,b_2=2) and F(a_3=0, b_3=1).` Now

`Area (ΔDEF)= 1/2 [a_1(b_2-b_2)+a_2(b_3-b_1)+a_3(b_1-b_2)]`

`=1/2 [1(2-1)+1(1-0)+0(0-2)]`

`=1/2[1+1+0]=1 `sq. units

So, the area of the triangle ΔDEF is 1 sq. unit.

Hence, ΔABC : ΔDEF = 4:1.