How do you verify that parallelogram ABCD with vertices A$( - 5, - 1)$,B$( - 9,6)$,C$( - 1,5)$ and D$(3, - 2)$ is a rhombus by showing that it is a parallelogram with a perpendicular diagonal?

Hint: First we know the slope of a parallelogram. To find the slope of a parallelogram. To find the slope of ${m_{AB}},{m_{CD}},{m_{BC}},{m_{AD}}$. The formula is,
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
If \[AB\parallel CD,{\text{ }}BC\parallel AD\]. It is a parallelogram.
And then we find the midpoint of AC and BD.
If AC and BD are midpoints, then it’s a rhombus.

Complete step-by-step solution:

The given information is that parallelogram ABCD with vertices A$( - 5, - 1)$,B$( - 9,6)$,C$( - 1,5)$ and D$(3, - 2)$.
If \[AB\parallel CD,{\text{ }}BC\parallel AD\]. It is a parallelogram.
If the slope of \[ABCD,{\text{ }}BCAD\] then it is a parallelogram
Now the slope of parallelogram formula is,
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Let AB vertices $( - 5, - 1)$ and $( - 9,6)$, now find the slope of AB
$\Rightarrow$${m_{AB}} = \dfrac{{6 + 1}}{{ - 9 + 5}}$
Add numerator and subtract the denominator,
$\Rightarrow$${m_{AB}} = - \dfrac{7}{4}$
Let CD vertices $( - 1,5)$and $(3, - 2)$, now find the slope of CD
$\Rightarrow$${m_{CD}} = \dfrac{{ - 2 - 5}}{{3 + 1}}$
Add numerator and the denominator
$\Rightarrow$${m_{CD}} = - \dfrac{7}{4}$
Let BC vertices $( - 9,6)$and $( - 1,5)$, now find the slope of BC
$\Rightarrow$${m_{BC}} = \dfrac{{5 - 6}}{{ - 1 + 9}}$
Subtract numerator and denominator
$\Rightarrow$${m_{BC}} = - \dfrac{1}{8}$
Let AD vertices $( - 5, - 1)$and $(3, - 2)$, now find the slope of AD
$\Rightarrow$${m_{AD}} = \dfrac{{ - 2 + 1}}{{3 + 5}}$
Subtract numerator and add the denominator
$\Rightarrow$${m_{AD}} = - \dfrac{1}{8}$
Hence \[AB\parallel CD,{\text{ }}BC\parallel AD\] and ABCD is parallelogram.
$\Rightarrow$Slope of AC ${m_{AC}} = \dfrac{{5 + 1}}{{ - 1 + 5}} = \dfrac{6}{4} = \dfrac{3}{2}$
$\Rightarrow$Slope of BD ${m_{BD}} = \dfrac{{6 + 2}}{{ - 9 - 3}} = \dfrac{8}{{ - 12}} = - \dfrac{2}{3}$
$\Rightarrow$${m_{AC}} = - \dfrac{1}{{{m_{BD}}}}$
Hence diagonals intersect each other at right angles.
It can be a square or a rhombus or a kite.
It’s not a square since AB is not perpendicular AD.
It can be a rhombus.
If the intersection point of the diagonals is their midpoint, then it’s a rhombus, else it’s a kite.
Let’s find the midpoint $(E)$ of AC and BD to prove it’s a rhombus.
$E = \dfrac{{A( - 5, - 1) + C( - 1,5)}}{2} = ( - 3,2)$
Also $E = \dfrac{{B( - 9,6) + D(3, - 2)}}{2} = ( - 3,2)$
Hence it’s a rhombus.

Note: Properties of a rhombus:
Opposite sides are parallel.
In a parallelogram, the opposite are equal.
Let, $ABCD$ is a parallelogram
Therefore, $AB = CD$ and $AD = BC$

Opposite angles are equal.
In a parallelogram, the opposite angles are equal.
Let, ABCD is a parallelogram
Therefore, $\angle A = \angle C$ and $\angle B = \angle D$

Diagonals bisect each other at right angles.

In a parallelogram, the diagonal bisects each other.
$AC$ and $BD$ are the diagonals of $ABCD$ bisecting each other at point $E$
$AE = EC$ and $EB = ED$