Question

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

Answer 1

Solve for the required area and ratio

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(0,-1\right) \\ B\left(x_2,y_2\right)=\left(2,1\right) \\ C\left(x_3,y_3\right)=\left(0,3\right) \end{gathered}$$

Step 1: Solve for mid-points of triangle

Midpoint of two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Therefore, midpoint of :

$$\begin{gathered} AB,D=\left(\frac{0+2}{2},\frac{-1+1}{2}\right)=\left(1,0\right) \\ AC,E=\left(\frac{0+0}{2},\frac{-1+3}{2}\right)=\left(0,1\right) \\ BC,F=\left(\frac{2+0}{2},\frac{1+3}{2}\right)=\left(1,2\right) \end{gathered}$$

Step 2: Solve for areas of triangles

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

$\therefore$Area of $△ABC=\frac{1}{2}\left|0\left(1-3\right)+2\left(3+1\right)+0\left(-1-1\right)\right|$

$$\begin{gathered} =\frac{1}{2}\left|0+8+0\right| \\ =4sq.units \end{gathered}$$

$\therefore$Area of $△DEF=\frac{1}{2}\left|1\left(1-2\right)+0\left(2+0\right)+1\left(0-1\right)\right|$

$$\begin{gathered} =\frac{1}{2}\left|-1+0-1\right| \\ =1sq.unit \end{gathered}$$

Hence, area of the triangle formed by midpoints of vertices $(0,-1),(2,1)$ and $\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{}\mathrm{3}\mathrm{)}$ is $1sq.unit$ and ratio of area of the triangle formed by midpoints to the area of the given triangle is $1:4$