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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.
Given:
The vertices of a triangle are $(0, -1), (2, 1)$ and $(0, 3)$.
To do:
We have to find the area of the triangle formed by joining the mid-points of the sides of the triangle and the ratio of this area to the area of the given triangle.
Solution:
Let $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are the mid-points of the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ respectively.
This implies,
Coordinates of the point $ \mathrm{D}=\frac{0+2}{2}, \frac{-1+1}{2}$
$=(1,0)$
Coordinates of the point $\mathrm{E}=\frac{2+0}{2}, \frac{1+3}{2}$
$=(1,2)$
Coordinates of the point $\mathrm{F}=\frac{0+0}{2}, \frac{3-1}{2}$
$=(0,1)$
Area of the triangle $\mathrm{DEF}=\frac{1}{2}[1(2-1)+1(1-0)+0(0-2)]$
$=\frac{1}{2}[1+1+0]$
$=\frac{2}{2}$
$=1$ sq units.
Area of triangle $\mathrm{ABC}=\frac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]$
$=\frac{1}{2}[0+8+0]$
$=4$ sq units
The ratio of areas of $\triangle \mathrm{DEF}$ and $\triangle \mathrm{ABC}$ is $1: 4$
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