Question

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.

Answer 1

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.