A(7, -3), B(5,3) and C(3,-1) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.

#### Solution

The vertices of the triangle are A(7, -3), B(5,3) and C(3,-1)

`"Coordinates of" D = ((5+3)/2,(3-1)/2) = (4,1)`

For the area of the triangle ADC, let

`A (x_1,y_1)=A(7,-3), D(x_2,y_2) =D(4,1) and C (x_3,y_3) = C(3,-1)`. Then

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

`=1/2 [7(1+1)+4(-1+3)+3(-3-1)]`

`=1/2[14+8-12}=5` sq. unit

Now, for the area of triangle ABD, let

`A(x_1,y_1) = A(7,-3), B(x_2,y_2) = B(5,3) and D (x_3,y_3) = D (4,1). `Then

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

`=1/2 [7(3-1)+5(1+3)+4(-3-3)]`

`=1/2[14+20-24] = 5` sq. unit

Thus, Area (ΔADC) = Area (ΔABD) = 5. sq units

Hence, AD divides ΔABC into two triangles of equal areas.