Find the area of triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

#### Solution

Let ABC be the triangle such that A (2, 1), B (4, 3) and C (2, 5) are the vertices of the triangle.

Let P, Q and R be the mid-points of sides AB, BC and CA respectively of ΔABC.

`\text{Coordination of the mid-point of points}(x_1,y_2) and (x_2,y_2)=((x_1+x_2)/2, (y_1+y_2)/2)`

`\text{Coordination of P}`=`((4+2)/2, (3+1)/2)=(3,2)`

`\text{Coordination of Q}`=`((4+2)/2, (3+5)/2)=(3,4)`

`\text{Coordination of Q}`=`((2+2)/2, (5+1)/2)=(2,3)`

The area of the triangle whose vertices are (*x*_{1} , *y*_{1} ), (*x*_{2} , *y*_{2} ) and (*x*_{3} , *y*_{3} ) is given as follows:

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

∴Area of ΔPQR

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

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

`=1/2[3+3-4]`

`=2/2`

`=1 sq.unit`

Thus, the area of the triangle formed by joining the mid-points of the sides of the given triangle

is 1 sq unit.