in triangle ABC the coordinates of vertex A are (0,-1) D(1,0) and E(0,1) respectively the mid points of sides AB and AC . if F is the mid point of side BC , find the area of triangle DEF. 

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Please find below the solution to the asked query:

We form our diagram from given information , As :


We know formula for coordinate of mid point ( x , y ) = $\left(\frac{x_1 + x_2}{2} , \frac{{\displaystyle y_1 + y_2}}{{\displaystyle 2}}\right)$

As D is mid point of AB , So x  =  1 , y  = 0  and x₁ = 0, x₂ =  x and  y₁ = - 1 , y₂ =  y  , ( As we assume coordinate of B(x,y ) ), So

$$\begin{gathered} \left(1 , 0 \right) = \left(\frac{0 + x}{2} , \frac{{\displaystyle - 1 +y}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(1 , 0 \right) = \left(\frac{x}{2} , \frac{{\displaystyle - 1 +y}}{{\displaystyle 2}}\right) , \mathrm{So} \\ \frac{x}{2} = 1 \\ \Rightarrow x = 2 \\ \mathrm{And} \\ \frac{{\displaystyle - 1 +y}}{{\displaystyle 2}} = 0 \\ \Rightarrow - 1 + y = 0 \\ \Rightarrow y = 1 \end{gathered}$$
So, Coordinate of B ( 2 , 1 )

Similarly , as E is mid point of AC , So x  =  0 , y  = 1  and x₁ = 0, x₂ =  m and  y₁ = - 1 , y₂ =  m  , ( As we assume coordinate of C (m,n ) ), So

$$\begin{gathered} \left(0 , 1 \right) = \left(\frac{0 + m}{2} , \frac{{\displaystyle - 1 +n}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(0 , 1 \right) = \left(\frac{m}{2} , \frac{{\displaystyle - 1 +n}}{{\displaystyle 2}}\right) , \mathrm{So} \\ \frac{m}{2} = 0 \\ \Rightarrow m = 0 \\ \mathrm{And} \\ \frac{{\displaystyle - 1 +n}}{{\displaystyle 2}} = 1 \\ \Rightarrow - 1 + n = 2 \\ \Rightarrow n = 3 \end{gathered}$$
So, Coordinate of C ( 0 , 3 )

We assume coordinate of F ( p,q ), So x₁ = 2, x₂ =  0 and  y₁ = 1 , y₂ = 3  , As F is mid point of BC

$$\begin{gathered} \left(p , q \right) = \left(\frac{2 + 0}{2} , \frac{{\displaystyle 1 +3}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(p , q \right) = \left(\frac{2}{2} , \frac{{\displaystyle 4}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(p , q \right) = \left(1 , 2\right) \end{gathered}$$

We know area of triangle from given three points  :

Area  = $\frac{1}{2}\left| x_1\left(y_2 - y_{3 }\right) + x_2\left(y_3 - y_{1 }\right) + x_3\left(y_1 - y_{2 }\right)\right|\hspace{0.17em}$

To find area of triangle ABC , Here x₁ = 0 , x₂ =  2 , x₃ = 0  and  y₁ = - 1 , y₂ =  1 , y₃ = 3

So,

Area of triangle ABC  = $\frac{1}{2}\left|0\left( 1 - 3\right) + 2\left( 3 - \left( - 1\right)\right) + 0\left( -1 - 1\right)\right|\hspace{0.17em} =\frac{1}{2}\left|0\left(- 2\right) + 2\left( 3 + 1\right) + 0\left( -2\right)\right|\hspace{0.17em} =\frac{1}{2}\left|0 + 2\left( 4\right) +0\right|=\frac{1}{2}\times 8$ = 4 unit square

And

To find area of triangle DEF , Here x₁ = 1 , x₂ =  0 , x₃ = 1  and  y₁ = 0 , y₂ =  1 , y₃ = 2

So,

Area of triangle DEF  = $\frac{1}{2}\left|1\left( 1 - 2\right) + 0\left( 2 -0\right) + 1\left( 0 - 1\right)\right|\hspace{0.17em} =\frac{1}{2}\left|1\left(- 1\right) + 0\left( 2\right) + 1\left( -1\right)\right|\hspace{0.17em} =\frac{1}{2}\left| -1 +0 +-1\right|=\frac{1}{2}\left| -2\right|=\frac{1}{2}\times 2$ = 1 unit square


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