Question

Find the area of the triangle formed by joining the mid-points of the sides of the triangle $ABC$ whose vertices are $A(0,-1),B(2,1)$ and $C(0,3)$. Find the ratio of this area of the triangle $ABC$.

Answer 1

Let the vertices of the triangle be $A(0,-1),B(2,1),C(0,3)$.
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by
D = $(0+2/2$ , $-1+$$\frac{1}{2}$) $=$ $(1,0)$
E = ($0+$$\frac{0}{2}$, $-3$ -$\frac{1}{2}$) $=$ $(0,1)$
F = $(2+0/2$ , $1$ +$\frac{3}{2}$) $=$ $(1,2)$

Area of a triangle $=$ $\frac{1}{2}$ $x_1(y_2-y_3)$ +$x_2(y_3-y_1)$+ $x_3(y_1-y_2)$

Area of $△DEF$ $=$ $\frac{1}{2}$ $\left[1(2-1)+1(1-0)+0(0-2)\right]$
$=$ $\frac{1}{2}$ $(1+1)$ $=1$ square units

Area of $△ABC$ $=$ $\frac{1}{2}$ $\left[0\right(1-3)+23-(-1)+0(-1-1\left)\right]$
= $\frac{1}{2}$ $8$ $=$ $4$ square units
Therefore, the required ratio is $1:4$.