Answer: Showed.
Step-by-step explanation: Given vertices of parallelogram abcd are a(-5,-1), b(-9,6), c(-1,5) and d(3,-2).
Since it is a parallelogram, so opposite sides must be equal. That is,
ab = cd, bc = ad.
We are to show abcd is a rhombus by showing that the diagonals ac and bd are perpendicular to each other.
Now,
[tex]ab=\sqrt{(-5+9)^2+(-1-6)^2}=\sqrt{16+49}=\sqrt{65},\\\\bc= \sqrt{(-9+1)^2+(6-5)^2}=\sqrt{64+1}=\sqrt{65}.[/tex]
So, ab = bc = cd = da, with proves that all the sides are equal.
Now, slope of diagonal ac will be
[tex]S_{ac}=\dfrac{5+1}{-1+5}=\dfrac{3}{2},[/tex]
and the slope of bd will be
[tex]S_{bd}=\dfrac{-2-6}{3+9}=-\dfrac{2}{3}.[/tex]
Therefore,
[tex]S_{ac}\times S_{bd}=-1.[/tex]
This shows that the diagonals ac and bd are perpendicular.
Thus, the parallelogram abcd is a rhombus.
