Coordinates of a point, components of a vector and midpoint of a segment

If we draw the sides of the rhombus to see the figure better

If now we remember that a rhombus is a figure with 4 equal sides and equal opposite angles we have $\overrightarrow{CA}=\overrightarrow{BD}$ and $\overrightarrow{CB}=\overrightarrow{AD}$.

Therefore, if we apply vector $\overrightarrow{CA}$ to point $B$, we will obtain point $D$, and similarly we can apply the vector $\overrightarrow{CB}$ to point $A$ to obtain point $D$.

We begin by calculating the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$:

$$\overrightarrow{CA}=A-C=(1,-2)-(7,3)=(1-7,-2-3)=(-6,-5)$$

$$\overrightarrow{CB}=B-C=(2,-3)-(7,3)=(2-7,-3-3)=(-5,-6)$$

If now we apply these vectors to the points $B$ and $A$ respectively we obtain:

$$D=\overrightarrow{CA}+B=(-6,-5)+(2,-3)=(-4,-8)$$

$$D=\overrightarrow{CB}+A=(-5,-6)+(1,-2)=(-4,-8)$$

To find the coordinates of point $E$, we can do so, for example, by finding the coordinates of the average point of the segments $CD$ or $AB$.

$$E=\dfrac{A+B}{2}=\Big(\dfrac{a_1+b_1}{2},\dfrac{a_2+b_2}{2} \Big)=\Big(\dfrac{1+2}{2},\dfrac{-2-3}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-5}{2}\Big)$$

$$E=\dfrac{C+D}{2}=\Big(\dfrac{c_1+d_1}{2},\dfrac{c_2+d_2}{2} \Big)=\Big(\dfrac{7-4}{2},\dfrac{3-8}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-5}{2}\Big)$$