check the picture below
now, we know is a parallelogram, so the diagonals will bisect
looking at the picture, the interior angles are not right-angles, and thus is not a rectangle, and is not a square either, due to the same reason
is it a rhombus? well, a rhombus, is a parallelogram, that's "slanted" per se, but regardless of how slanted it may be, the sides are all equal
now, we don't need to check the length of both pairs, just one of the segments of each pair, let's check only then JK and KL, the ones in red in the picture, since each pair of segments are twin, if JK = KL, then that means all sides are equal
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
J&({{-7 }}\quad ,&{{ -2}})\quad
% (c,d)
K&({{ 0}}\quad ,&{{ 4}})\\\\
K&({{ 0}}\quad ,&{{ 4}})\quad
% (c,d)
L&({{ 9}}\quad ,&{{ 2}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
JK=\sqrt{[0-(-7)]^2+[4-(-2)]^2}\implies JK=\sqrt{(0+7)^2+(4+2)^2}
\\\\\\
JK=\sqrt{7^2+6^2}
\\\\\\
KL=\sqrt{(9-0)^2+(2-4)^2}\implies KL=\sqrt{9^2+(-2)^2}[/tex]
now... another characteristic of a rhombus is, the diagonals, meet at right-angle, that means KM ⟂ JL
and that simply means, if you get the slope of KM and the slope of JL, the product of their slope is -1
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
K&({{ 0}}\quad ,&{{ 4}})\quad
% (c,d)
M&({{ 2}}\quad ,&{{ -4}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-4-4}{2-0}[/tex]
[tex]\bf -------------------------------\\\\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
J&({{ -7}}\quad ,&{{ -2}})\quad
% (c,d)
L&({{ 9}}\quad ,&{{ 2}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{2-(-2)}{9-(-7)}\implies \cfrac{2+2}{9+7}[/tex]
then multiply both slopes, if their product is -1, that means, they're perpendicular to each other, and it IS a rhombus only, then.
