Find area of the triangle formed by joining the midpoints of the sides of the triangle
whose vertices are A(2, 1), B(4, 3) and C(2, 5).
The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
Coordinates of midpoint of $A B=P\left(x_{1}, y_{1}\right)=\left(\frac{2+4}{2}, \frac{1+3}{2}\right)=(3,2)$
Coordinates of midpoint of $B C=Q\left(x_{2}, y_{2}\right)=\left(\frac{4+2}{2}, \frac{3+5}{2}\right)=(3,4)$
Coordinates of midpoint of $A C=R\left(x_{3}, y_{3}\right)=\left(\frac{2+2}{2}, \frac{1+5}{2}\right)=(2,3)$
Now
Area of $\Delta P Q R=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
$=\frac{1}{2}[3(4-3)+3(3-2)+2(2-4)]$
$=\frac{1}{2}[3+3-4]=1$ sq. unit
Hence, the area of the required triangle is 1 sq. unit.
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