Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3)

Question:

Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3) and (−3,−2) is a rhombus.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In a rhombus all the sides are equal in length.

Here the four points are (21), B (3,  4), (23) and (32).

First let us check if all the four sides are equal.

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

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

 

$=\sqrt{1+25}$

$A B=\sqrt{26}$

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

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

 

$=\sqrt{25+1}$

$B C=\sqrt{26}$

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

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

 

$=\sqrt{25+1}$

$C D=\sqrt{26}$

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

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

 

$=\sqrt{25+1}$

$A D=\sqrt{26}$

Here, we see that all the sides are equal, so it has to be a rhombus.

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