Question

Find the area of $∆ABC$ 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. [CBSE 2014]

Answer 1

Let A(x₁ = 0, y₁ = −1), B(x₂ = 2, y₂ = 1) and C(x₃ = 0, y₃ = 3) be the given points. Then
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[0\left(1-3\right)+2\left(3+1\right)+0\left(-1-1\right)\right] \\ =\frac{1}{2}\times 8=4\mathrm{sq}.\mathrm{units} \end{gathered}$$
So, the area of the triangle $∆ABC$ is 4 sq. units.
Let D(a₁, b₁), E(a₂, b₂) and F(a₃, b₃) be the midpoints of AB, BC and AC respectively. Then
$$\begin{gathered} a_1=\frac{0+2}{2}=1b_1=\frac{-1+1}{2}=0 \\ {a}_2=\frac{2+0}{2}=1b_2=\frac{1+3}{2}=2 \\ {a}_3=\frac{0+0}{2}=0b_3=\frac{-1+3}{2}=1 \end{gathered}$$
Thus, the coordinates of D, E and F are D(a₁ = 1, b₁ = 0), E(a₂ = 1, b₂ = 2) and F(a₃ = 0, b₃ = 1). Now
$$\begin{gathered} \mathrm{Area}\left(∆DEF\right)=\frac{1}{2}\left[a_1\left(b_2-b_3\right)+a_2\left(b_3-b_1\right)+a_3\left(b_1-b_2\right)\right] \\ =\frac{1}{2}\left[1\left(2-1\right)+1\left(1-0\right)+0\left(0-2\right)\right] \\ =\frac{1}{2}\left[1+1+0\right]=1\mathrm{sq}.\mathrm{unit} \end{gathered}$$
So, the area of the triangle $∆DEF$ is 1 sq. unit.
Hence, $∆ABC:∆DEF=4:1$.