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$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.

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Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. –  Arturo Magidin Mar 12 '12 at 2:11
    
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@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
    
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1 Answer 1

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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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