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Coordinate Geometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsCoordinate Geometry3 Marks
Question
In the given figure, $\text{ABC}$ is a triangle coordinates of whose vertex $A$ are $(0,-1) D$ and E respectively are the mid $-$ points of the sides $A B$ and $A C$ and their coordinates are $(1,0)$ and $(0,1)$ respectively. If $F$ is the mid $-$ point of $B C$, find the areas of $\triangle A B C$ and $\triangle D E F$.
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Answer

Let the coordinates of $B$ and $C$ be $\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$.
Since, $D$ is the mid $-$ point of $A B$. So,
$(1,0)=\left(\frac{x_2+0}{2}, \frac{y_2-1}{2}\right)$
$a \Rightarrow 1=\frac{x_2+0}{2}$ and $ 0=\frac{y_2-1}{2}$
$\Rightarrow 1=\frac{x_2}{2} $ and $ 0=\frac{y_2-1}{2}$
$\Rightarrow x_2=2 $ and $y_2=1$
Thus, the coordinates of $B$ are $(2,1)$.
Similarly, $E$ is the mid $-$ point of $A C$. So,
$(0,1)=\left(\frac{x_3+0}{2}, \frac{y_3-1}{2}\right)$
$\Rightarrow 0=\frac{x_3+0}{2}$ and $1=\frac{y_3-1}{2}$
$\Rightarrow 0=\frac{x_3}{2}$ and $1=\frac{y_3-1}{2} $
$\Rightarrow x_3=0$ and $y_3=3$
Thus, the coordinates of $C$ are $(0,3)$.
We know that $F$ is the mid $-$ point of $B C$.
So, its coordinates are
$\left(\frac{2+0}{2}, \frac{1+3}{2}\right)=(1,2)$
Now we know that the area of the triangle is given by $-$
$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
So, area of $\triangle A B C$ is
$=\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|$
$=\frac{1}{2} \times 8=4$ square unit 
Area of the is $\triangle D E F$
$=\frac{1}{2}|1(1-2)+0(2-0)+1(0-1)|$
$=\frac{1}{2} \times|-2|=\frac{1}{2} \times 2=1$ square unit
Hence, the area of $\triangle A B C$ is $4$ and the area of $\triangle D E F$ is $1$ square unit.

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