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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Updated on: 10-Oct-2022

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