if vertices of paralleogram abcd are a(4,-1) b(5,3) c(2,5) d(1,1) if E is midpoint of AB find the coordinates of e and show that ar ABCD =2 arCDE

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Please find below the solution to the asked query:

We form our diagram from given information , As :



Here we assume coordinate of E ( x , y  )
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 E is mid point of AB , x₁ = 4, x₂ =  5 and  y₁ = - 1 , y₂ = 3  , ( And we assume coordinate of E (x,y ) ), So

$$\begin{gathered} \left(x , y \right) = \left(\frac{4 + 5}{2} , \frac{{\displaystyle - 1 +3}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(x , y \right) = \left(\frac{9}{2} , \frac{{\displaystyle 2}}{{\displaystyle 2}}\right) \\ \Rightarrow \left(x , y \right) = \left( 4.5 , 1\right) \end{gathered}$$

So, Coordinate of E ( 4.5 , 1 )                                                ( Ans )

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 ABD , Here x₁ = 4 , x₂ =  5 , x₃ = 1  and  y₁ = - 1 , y₂ =  3 , y₃ = 1

So,

Area of triangle ABD  = $\frac{1}{2}\left|4\left( 3 - 1\right) + 5\left( 1 - \left( - 1\right)\right) + 1\left( -1 - 3\right)\right|\hspace{0.17em} =\frac{1}{2}\left|4\left( 2\right) + 5\left( 1 + 1\right) + 1\left( - 4\right)\right|=\frac{1}{2}\left| 8 + 5 \left(2\right) - 4\right|=\frac{1}{2}\left| 8 + 10 - 4\right| = \frac{1}{2}\times 14$ = 7 unit square


We know diagonal of parallelogram divide it in two equal parts , So

Area of parallelogram ABCD = 2 ( Area of triangle ABD ) , So

Area of parallelogram ABCD = 2 ( 7 ) = 14 square unit

And

To find area of triangle CDE , Here x₁ = 2 , x₂ =  1 , x₃ = 4.5  and  y₁ = 5 , y₂ =  1 , y₃ = 1

So,

Area of triangle CDE  = $\frac{1}{2}\left|2\left( 1 - 1\right) + 1\left( 1 -5\right) + 4.5\left( 5 - 1\right)\right|\hspace{0.17em} =\frac{1}{2}\left|2\left( 0\right) + 1\left( - 4 \right) + 4.5\left( 4 \right)\right|\hspace{0.17em} =\frac{1}{2}\left| 0 -4 +18 \right|=\frac{1}{2}\times 14$ = 7 unit square


Now from area of parallelogram ABCD and triangle CDE we can say :

Area of ABCD  = 2 Area of triangle CDE                                                             ( Hence proved )



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