Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

The points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a quadrilateral. We need to show that this quadrilateral is a rhombus.

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Distance between points A(-3, 2) and B(-5, -5):

$= \sqrt{((-5) - (-3))^2 + ((-5) - 2)^2}\\[1em] = \sqrt{(-2)^2 + (-7)^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

Distance between points B(-5, -5) and C(2, -3):

$= \sqrt{(2 - (-5))^2 + ((-3) - (-5))^2}\\[1em] = \sqrt{7^2 + 2^2}\\[1em] = \sqrt{49 + 4}\\[1em] = \sqrt{53}$

Distance between points C(2, -3) and D(4, 4):

$= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

Distance between points D(4, 4) and A(-3, 2):

$= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

AB = BC = CD = DA = $\sqrt{53}$

Since all sides are equal, the quadrilateral is a rhombus.

Hence, the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.