# Given two vertices, how to find the other two vertices of a rhombus?

$A\;(-3,-4)$ 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 $D$?

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Well, I'm in a bit of a hurry so didn't have time to view the site etiquette; but thanks for letting me know anyways. (and I don't mean this sarcastically) –  Deniz Mar 12 '12 at 2:31
@Arturo, your comment is very well written; would it be okay if I copy it verbatim to other questions by new users that have the same issues? –  Rahul Mar 12 '12 at 5:31
@Deniz:I have edited your question to less imperative form. Please check. –  Quixotic Mar 12 '12 at 7:02
@RahulNarain: Since that is what I did in the first place (copy it off someone else) of course it's fine. (-: –  Arturo Magidin Mar 12 '12 at 14:54

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 $D=(0,1 )$.

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