Show that A (−3, 2), B (−5, −5), C (2,−3), and

Let A (−3, 2); B (−5,−5); C (2,−3) and D (4, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rhombus.

So we should find the lengths of sides of quadrilateral ABCD.

$\mathrm{AB}=\sqrt{(-5+3)^{2}+(-5-2)^{2}}$

$=\sqrt{4+49}$

$=\sqrt{53}$

$B C=\sqrt{(2+5)^{2}+(-3+5)^{2}}$

$=\sqrt{4+49}$

$=\sqrt{53}$

$\mathrm{CD}=\sqrt{(4-2)^{2}+(4+3)^{2}}$

$=\sqrt{4+49}$

$=\sqrt{53}$

$\mathrm{AD}=\sqrt{(4+3)^{2}+(4-2)^{2}}$

$=\sqrt{4+49}$

$=\sqrt{53}$

All the sides of quadrilateral are equal. Hence ABCD is a rhombus.