Show that the points $ (1,-2),(2,3),(-3,2) $ and $ (-4,-3) $ are the vertices of a rhombus.

Given:

Given points are \( (1,-2),(2,3),(-3,2) \) and \( (-4,-3) \).

To do:

We have to show that the \( (1,-2),(2,3),(-3,2) \) and \( (-4,-3) \) are the vertices of a rhombus.

Solution:

Let \( \mathrm{ABCD} \) is a quadrilateral whose vertices are \( \mathrm{A}(1,-2), \mathrm{B}(2,3), \mathrm{C}(-3,2) \) and \( \mathrm{D}(-4,-3) \).

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,

 \( A B=\sqrt{(2-1)^{2}+(3+2)^{2}} \)

\( =\sqrt{(1)^{2}+(5)^{2}} \)

\( =\sqrt{1+25} \)

\( =\sqrt{26} \)

Similarly,

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

\( =\sqrt{(-5)^{2}+(-1)^{2}} \)

\( =\sqrt{25+1} \)

\( =\sqrt{26} \)

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

\( =\sqrt{(-1)^{2}+(-5)^{2}} \)

\( =\sqrt{1+25} \)

\( =\sqrt{26} \)

\( D A=\sqrt{(-4-1)^{2}+(-3+2)^{2}} \)

\( =\sqrt{(-5)^{2}+(-1)^{2}} \)

\( =\sqrt{25+1} \)

\( =\sqrt{26} \)

Diagonal \( \mathrm{AC}=\sqrt{(-3-1)^{2}+(2+2)^{2}} \)

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

\( =\sqrt{16+16} \)

\( =\sqrt{32} \)

\( =\sqrt{16 \times 2} \)

\( =4 \sqrt{2} \)

Diagonal \( \mathrm{BD}=\sqrt{(-4-2)^{2}+(-3-3)^{2}} \)

\( =\sqrt{(-6)^{2}+(-6)^{2}} \)

\( =\sqrt{36+36} \)

\( =\sqrt{72} \)

\( =\sqrt{36 \times 2} \)

\( =6 \sqrt{2} \)

Here,

$A B=B C=C D=D A=\sqrt{26}$

Sides are equal but diagonals are not equal.

Therefore, the given points are the vertices of a rhombus.