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

Given:

Given points are $(-3, 2), (-5, -5), (2, -3)$ and $(4, 4)$.

To do:

We have to show that the $(-3, 2), (-5, -5), (2, -3)$ and $(4, 4)$ are the vertices of a rhombus.

Solution:

Let \( \mathrm{ABCD} \) is a quadrilateral whose vertices are \( \mathrm{A}(-3,2), \mathrm{B}(-5,-5), \mathrm{C}(2,-3) \) and \( \mathrm{D}(4,4) \).

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

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

\( =\sqrt{(-2)^{2}+(-7)^{2}} \)

\( =\sqrt{4+49} \)

\( =\sqrt{53} \)

Similarly,

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

\( =\sqrt{(7)^{2}+(2)^{2}} \)

\( =\sqrt{49+4} \)

\( =\sqrt{53} \)

\( C D=\sqrt{(4-2)^{2}+(4+3)^{2}} \)

\( =\sqrt{(2)^{2}+(7)^{2}} \)

\( =\sqrt{4+49} \)

\( =\sqrt{53} \)

\( D A=\sqrt{(-3-4)^{2}+(2-4)^{2}} \)

\( =\sqrt{(-7)^{2}+(-2)^{2}} \)

\( =\sqrt{49+4} \)

\( =\sqrt{53} \)

Diagonal \( \mathrm{AC}=\sqrt{(2+3)^{2}+(-3-2)^{2}} \)

\( =\sqrt{(5)^{2}+(-5)^{2}} \)

\( =\sqrt{25+25} \)

\( =\sqrt{50} \)

\( =\sqrt{25 \times 2} \)

\( =5 \sqrt{2} \)

Diagonal \( \mathrm{BD}=\sqrt{(4+5)^{2}+(4+5)^{2}} \)

\( =\sqrt{(9)^{2}+(9)^{2}} \)

\( =\sqrt{81+81} \)

\( =\sqrt{162} \)

\( =\sqrt{81 \times 2} \)

\( =9 \sqrt{2} \)

Here,

$A B=B C=C D=D A=\sqrt{53}$

Sides are equal but diagonals are not equal.

Therefore, $A (-3, 2), B (-5, -5), C (2, -3)$ and $D (4, 4)$ are the vertices of a rhombus.