Question

Prove that the points $A(2,3),B(-2,2),C(-1,-2)$ and $D(3,-1)$ are the vertices of a square $ABCD$.

Answer 1

We have,

$A(x_1,y_1)=(2,3)$

$B(x_2,y_2)=(-2,2)$

$C(x_3,y_3)=(-1,-2)$

$D(x_4,y_4)=(3,-1)$

Now,

Using distance formula we get,

$AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

$\Rightarrow AB=\sqrt{{(2+2)}^2+{(3-2)}^2}$

$\Rightarrow AB=\sqrt{17}$

$BC=\sqrt{{(x_2-x_3)}^2+{(y_2-y_3)}^2}$

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

$\Rightarrow BC=\sqrt{17}$

$CD=\sqrt{{(x_3-x_4)}^2+{(y_3-y_4)}^2}$

$\Rightarrow CD=\sqrt{{(-1-3)}^2+{(-2+1)}^2}$

$\Rightarrow CD=\sqrt{17}$

$DA=\sqrt{{(x_4-x_1)}^2+{(y_4-y_1)}^2}$

$\Rightarrow DA=\sqrt{{(3-2)}^2+{(-1-3)}^2}$

$\Rightarrow DA=\sqrt{17}$

Now diagonal $AC=\sqrt{{(x_1-x_3)}^2+{(y_1-y_3)}^2}$

$\Rightarrow AC=\sqrt{{(2+1)}^2+{(3+2)}^2}$

$\Rightarrow AC=\sqrt{34}$

And diagonal $BD=\sqrt{{(x_2-x_4)}^2+{(y_2-y_3)}^4}$

$\Rightarrow BD=\sqrt{{(-2-3)}^2+{(2+1)}^2}$

$\Rightarrow BD=\sqrt{34}$

So,

Sides $AB=BC=CD=DA$ and diagonals $AC=BD$

Hence, the vertices are of a square $ABCD$.

Hence proved.