Given $$3$$ apexes of a rhombus $$ABCD$$, $$A = (1,-2)$$, $$B = (2,-3)$$, $$C = (7, 3)$$, find the coordinates of the apex $$D$$ and of the center $$E$$.
Development:
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 $$AB$$.
$$$E=\dfrac{A+B}{2}=\Big(\dfrac{a_1+b_1}{2},\dfrac{a_2+b_2}{2} \Big)=\Big(\dfrac{1+2}{2},\dfrac{-2-3}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-5}{2}\Big)$$$
$$$E=\dfrac{C+D}{2}=\Big(\dfrac{c_1+d_1}{2},\dfrac{c_2+d_2}{2} \Big)=\Big(\dfrac{7-4}{2},\dfrac{3-8}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-5}{2}\Big)$$$
Solution:
$$D=(-4,8)$$
$$E=\Big(\dfrac{3}{2},-\dfrac{5}{2}\Big)$$