Question

Show 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

Length OF SIDES-

$AB=\sqrt{\left[\right(2+2{)}^2+(3-2{)}^2]}$

$=\sqrt{[16+1]}$

$=\sqrt{17}$

$BC=\sqrt{\left[\right(-2+1{)}^2+(2+2{)}^2]}$

$=\sqrt{[1+16]}$

$=\sqrt{17}$

$CD=\sqrt{\left[\right(-1-3{)}^2+(-2+1{)}^2]}$

$=\sqrt{17}$

$DA=\sqrt{\left[\right(3-2{)}^2+(-1-3{)}^2]}$

$=\sqrt{17}$

here $AB=BC=CD=DA$

LENGTH OF DIAGONALS

$AC=\sqrt{\left[\right(2+1{)}^2+(3+2{)}^2]}$

$=\sqrt{[9+25]}$

$=\sqrt{34}$

$BD=\sqrt{\left[\right(-2-3{)}^2+(2+1{)}^2]}$

$=\sqrt{34}$

here $AC=BD$

hence, the given points are vertices of a square