Show that the points $A( 2,\ -1),\ B( 3,\ 4),\ C( -2,\ 3)$ and $D( -3,\ -2)$ are the vertices of a rhombus.
Given: Points $A( 2,\ -1),\ B( 3,\ 4),\ C( -2,\ 3)$ and $D( -3,\ -2)$
To do: To show that the given points are the vertices of a rhombus.
Solution:
$AB=\sqrt{(3-2)^2+(4+1)^2}=\sqrt{1+25}=\sqrt{26}$
$BC=\sqrt{(-2-3)^2+(3-4)^2}=\sqrt{25+1}=\sqrt{26}$
$CD=\sqrt{(-3+2)^2+(-2-3)^2}=\sqrt{1+25}=\sqrt{26}$
$AD=\sqrt{-3-2)^2+(-2+1)^2}=\sqrt{25+1}=\sqrt{26}$
$AC=\sqrt{(-2-2)^2+(3+1)^2}=\sqrt{16+16}=\sqrt{32}$
$BD=\sqrt{(-3-3)^2+(-2-4)^2}=\sqrt{36+36}=\sqrt{72}$
Here, we find that sides $AB=BC=CD=AD$, but diagonals $AC$ and $BD$ are not equal.
Therefore, it is a rhombus.
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