Why do rhombus diagonals intersect at right angles?

Another way to say that the slopes are opposite reciprocals is to say that their product is $-1$.

$$\begin{align} \frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{-\sqrt{a^2-b^2}}{a-b} &=\frac{-(\sqrt{a^2-b^2})^2}{(a+b)(a-b)} \\ &=\frac{-(a^2-b^2)}{a^2-b^2} \\ &=-1 \end{align}$$

You don't have to work through square roots if you use the properties of the vector dot product and the parallelogram law to construct the rhombus.

I.e one of the rhombus's diagonals can be identified with $\mathbf{a + b}$ where $\mathbf{b}$ is a vector added head-to-tail to vector $\mathbf{a}$, according to the parallelogram law. Similarly the other diagonal is given by $\mathbf{b - a}$. The constraint for a rhombus is $\lVert \mathbf{a} \rVert^2 = \lVert \mathbf{b} \rVert^2$. Two vectors $\mathbf{u, v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$, and as $\mathbf{(a + b)} \, \mathbf{\cdot} \, \mathbf{(b - a)}$ = $ \lVert \mathbf{b} \rVert^2 - \lVert \mathbf{a} \rVert^2 = 0$, the two diagonals are perpendicular.

Hint: Multiply the slopes together and simplify.