Three vertices of the parallelogram are given and we need to find the coordinates of the fourth vertex and also the area of the parallelogram.

Let the coordinates of vertex 𝐷 be (𝑥, 𝑦). 

It is known that in a parallelogram, diagonals bisect each other. Therefore, 𝐸 acts as the midpoint of the diagonals.

On using midpoint formula, we get the coordinates of point 𝐸 as

$\begin{array}{l}\Rightarrow E=\left(\frac{3+(-6)}{2},\frac{(-4)+2}{2}\right)=\left(\frac{x+(-1)}{2},\frac{(-3)+y}{2}\right)\\ \Rightarrow E=\left(\frac{-3}{2},\frac{-2}{2}\right)=\left(\frac{x-1}{2},\frac{y-3}{2}\right)\end{array}$On considering the last two terms, we get

$\left(\frac{-3}{2},\frac{-2}{2}\right)=\left(\frac{x-1}{2},\frac{y-3}{2}\right)$

Therefore,

$\frac{𝑥-1}{2} = \frac{-3 }{2}$

⇒ 𝑥 − 1 =− 3 

⇒ 𝑥 =− 2 

Similarly,

$\begin{array}{l}\frac{y-3}{2}=\frac{-2}{2}\\ \Rightarrow y-3=-2\\ \Rightarrow y=1\end{array}$

Hence, the coordinates of vertex 𝐷 is (− 2, 1). 

It is known that the diagonal of the parallelogram divides it into two equal parts having the same area. Therefore, the diagonal 𝐴𝐶 divides the parallelogram 𝐴𝐵𝐶𝐷 into two equal triangles ∆𝐴𝐵𝐶 and ∆𝐴𝐷𝐶 

Also, if coordinates of points 𝑃, 𝑄 and 𝑅 are$(x_1 y_1), (x_2 y_2) and (x_3 y_3)$ respectively, then the area of the triangle formed by them is

$A=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_1\left(y_1-y_2\right)\right]$

$\text{ If we assume }P\left(x_1,y_1\right)=A(3,-4),Q\left(x_2,y_2\right)=B(-1,-3)\text{ and }R\left(x_3,y_3\right)=C$(− 6, 2), then on substituting the values in the above relation, we get

$A=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_1\left(y_1-y_2\right)\right]$

$\begin{array}{l}\Rightarrow A=\frac{1}{2}\left[3\right(-3-2)+(-1\left)\right(2-(-4))+(-6\left)\right(-4-(-3)\left)\right]\\ \Rightarrow A=\frac{1}{2}\left[3\right(-5)-1(6)+(-6\left)\right(-1\left)\right]\end{array}$$\begin{array}{l}\Rightarrow A=\frac{1}{2}[-15-6+6]\\ \Rightarrow A=\frac{-15}{2}\\ \Rightarrow A=\frac{15}{2}\text{ sq. units }\end{array}$

Hence, the area of the triangle ∆𝐴𝐵𝐶 is  $\frac{15}{2}\text{ sq. units. }$

Therefore, the area of parallelogram is double the area of the triangle, i.e., 15 𝑠𝑞. 𝑢𝑛𝑖𝑡.

                                                             (OR)

We need to prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides. Let the triangles be ∆𝐴𝐵𝐶 and ∆𝑃𝑄R.

Now, according to the question, ΔABC ~ ΔPQR and we need to prove that

$$\frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$$

On drawing AD⟂BC and PE⟂QR, we get the diagram as shown below

Since ΔABC ~ ΔPQR, it can be said that the corresponding angles are equal. 

Therefore, both the triangles are equiangular and hence their corresponding sides are proportional. It can be written as $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR} ...\left(i\right)$

Also, since

ΔABC ~ ΔPQR, it can be said thatΔADB ~ ΔPEQ.  Therefore, 

$\frac{AD}{PE}=\frac{AB}{PQ} .......\left(ii\right)$

Now, from the above two equations, 

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PE} ......\left(iii\right)$

Now, the ratio of areas of ΔABC  and ΔPQR  is given as 

$$\frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}=\frac{0.5\times BC\times AD}{0.5\times QR\times PE}$$$$\Rightarrow \frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}=\frac{BC}{QR}\times \frac{AD}{PE}$$

From the third equation, we get 

$$\Rightarrow \frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$$

Hence, proved.