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In $\triangle ABC$, the coordinates of vertex $A$ are $(0, -1)$ and $D (1, 0)$ and $E (0, 1)$ respectively the mid-points of the sides $AB$ and $AC$. If $F$ is the mid-point of side $BC$, find the area of $\triangle DEF$.
Given:
In $\triangle ABC$, the coordinates of vertex $A$ are $(0, -1)$ and $D (1, 0)$ and $E (0, 1)$ respectively the mid-points of the sides $AB$ and $AC$.
$F$ is the mid-point of side $BC$.
To do:
We have to find the area of $\triangle DEF$.
Solution:
Let $B (x_2, y_2), C (x_3, y_3)$ be the other two vertices of the $\triangle ABC$ and $F(h,k)$ be the mid-point of $BC$.
$D$ is the mid-point of $AB$.
This implies,
\( (\frac{0+x_{2}}{2}, \frac{-1+y_2}{2})=(1,0) \)
On comparing, we get,
\( \frac{x_2}{2}=1 \) and \( \frac{-1+y_2}{2}=0 \)
\( x_2=1(2) \) and \( -1+y_2=0(2) \)
\( x_2=2 \) and \( y_2=0+1=1 \)
Similarly,
\( E \) is the mid-point of \( A C \).
\( (\frac{0+x_{3}}{2}, \frac{-1+y_3}{2})=(0,1) \)
On comparing, we get,
\( \frac{x_3}{2}=0 \) and \( \frac{-1+y_3}{2}=1 \)
\( x_3=0(2) \) and \( -1+y_3=1(2) \)
\( x_3=0 \) and \( y_3=2+1=3 \)
\( \mathrm{F} \) is the mid-point of \( \mathrm{BC} \).
\( (\frac{x_3+x_{2}}{2}, \frac{y_3+y_2}{2})=(h,k) \)
On comparing, we get,
\( \frac{2+0}{2}=h \) and \( \frac{1+3}{2}=k \)
\( 2(h)=2 \) and \( 2(k)=4 \)
\( h=1 \) and \( k=2 \)
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( DEF=\frac{1}{2}[1(1-2)+0(2-0)+1(0-1)] \)
\( =\frac{1}{2}[1(-1)+0+1(-1)] \)
\( =\frac{1}{2}(-1-1) \)
\( =\frac{1}{2}(-2) \)
\( =1 \) sq. unit
The area of $\triangle DEF$ is $1$ sq. unit.
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