Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus.

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).

$A B=\sqrt{(-5+3)^{2}+(-5-2)^{2}}=\sqrt{(-2)^{2}+(-7)^{2}}=\sqrt{4+49}=\sqrt{53}$ units.

$B C=\sqrt{(2+5)^{2}+(-3+5)^{2}}=\sqrt{(7)^{2}+(2)^{2}}=\sqrt{49+4}=\sqrt{53}$ units.

$C D=\sqrt{(4-2)^{2}+(4+3)^{2}}=\sqrt{(2)^{2}+(7)^{2}}=\sqrt{4+49}=\sqrt{53}$ units.

$D A=\sqrt{(4+3)^{2}+(4-2)^{2}}=\sqrt{(7)^{2}+(2)^{2}}=\sqrt{49+4}=\sqrt{53}$ units.

Therefore, $A B=B C=C D=D A=\sqrt{53}$ units

$A C=\sqrt{(2+3)^{2}+(-3-2)^{2}}=\sqrt{(5)^{2}+(-5)^{2}}=\sqrt{25+25}=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2}$ units

$B D=\sqrt{(4+5)^{2}+(4+5)^{2}}=\sqrt{(9)^{2}+(9)^{2}}=\sqrt{81+81}=\sqrt{162}=\sqrt{81 \times 2}=9 \sqrt{2}$ units

Thus, diagonal $A C$ is not equal to diagonal $B D$.

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.

Area of a rhombus $=\frac{1}{2} \times$ (product of diagonals)

$=\frac{1}{2} \times(5 \sqrt{2}) \times(9 \sqrt{2})$

$=\frac{45(2)}{2}$

$=45$ square units