Question

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

Answer 1

Let $A(-3,2),B(-5,-5),C(2,-3)$ and $D(4,4)$

Now we are going to find distance between sides.

$\Rightarrow$ $AB=\sqrt{(-5+3{)}^2+(-5-2{)}^2}=\sqrt{(-2{)}^2+(-7{)}^2}=\sqrt{4+49}=\sqrt{53}$

$\Rightarrow$ $BC=\sqrt{(2+5{)}^2+(-3+5{)}^2}=\sqrt{(7{)}^2+(2{)}^2}=\sqrt{49+4}=\sqrt{53}$

$\Rightarrow$ $CD=\sqrt{(4-2{)}^2+(4+3{)}^2}=\sqrt{(2{)}^2+(7{)}^2}=\sqrt{4+49}=\sqrt{53}$

$\Rightarrow$ $DA=\sqrt{(-3-4{)}^2+(2-4{)}^2}=\sqrt{(-7{)}^2+(-2{)}^2}=\sqrt{49+4}=\sqrt{53}$

$\therefore$ $AB=BC=CD=DA$

We can see all sides are equal.

Now, we are going to find distance between diagonals.

$\Rightarrow$ $AC=\sqrt{(2+3{)}^2+(-3-2{)}^2}=\sqrt{(5{)}^2+(-5{)}^2}=\sqrt{25+25}=\sqrt{50}$

$\Rightarrow$ $BD=\sqrt{(4+5{)}^2+(4+5{)}^2}=\sqrt{(9{)}^2+(9{)}^2}=\sqrt{81+81}=\sqrt{162}$

$\therefore$ $AC\ne BD$

All sides are equal but diagonals are not equal, hence, it is a rhombus.