$A\;(-3,-4) $ and $ C \; (5,4)$ are the ends of the diagonal of a rhombus $ABCD$.
Given that the side BC has gradient $\frac{5}{3}$; How could we find the coordinates of $B$ and hence of $D$?
Context
In a rhombus, both diagonals intersect at their midpoint. Therefore, it suffices to find $B$, from where $D$ can be found by using $B+D=A+C$. But how to find $B$? The slope of BC is known, but not its length.
Let $B=(x,y)$ be the vertex below the given diagonal. Since $B$ is equidistant to $A$ and $C$: $$ (x-5)^2+(y-4)^2=(x+3)^2+(y+4)^2. $$ Simplifying the above yields $$\tag{1} 1=x+y $$
Since $BC$ has gradient $5/3$ $$ {y-4\over x-5}={5\over3}; $$ whence $$\tag{2}3y-5x=-13.$$
By $(1)$, we have $y=1-x$. Substituting into $(2)$ gives $ 3(1-x)-5x=-13$. This gives $x=2$; and, from $(1)$, $y=-1$.
So $B=(2,-1)$.
Since the given diagonal has slope 1, and since $B$ is 5 units to the right and 3 units up from $A$, the forth vertex is 5 units up and 3 to the right of $A$, So $D=(0,1 )$.