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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 $D$?


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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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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I got the answer in a very much more long winded way by using simultaneous equations. I don't understand the first part of the solution. How is B equidistant to AC? – dagda1 Sep 19 '15 at 10:41
@dagda1 In a rhombus, all sides share a common length. $B$ and $A$ form one side of the given rhombus, $B$ and $C$ form another. – David Mitra Sep 19 '15 at 10:54

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