Question

Show that the point A (3,5) B (6,0) C (1,-3) D (-2,2) are the vertex of the square ABCD.

Answer 1

We have given points are

$A(x_1,y_1)=(3,5)$

$B(x_2,y_2)=(6,0)$

$C(x_3,y_3)=(1,-3)$

$D(x_4,y_4)=(-2,2)$

Then,

Distance between

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

$=\sqrt{{(3-6)}^2+{(5-0)}^2}$

$=\sqrt{9+25}$

$AB=\sqrt{34}$

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

$=\sqrt{{(6-1)}^2+{(0+3)}^2}$

$=\sqrt{25+9}$

$BC=\sqrt{34}$

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

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

$=\sqrt{9+25}$

$CD=\sqrt{34}$

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

$=\sqrt{{(-2-3)}^2+{(2-5)}^2}$

$=\sqrt{25+9}$

$DA=\sqrt{34}$

Then, $AB=BC=CD=DA$

All sides are equal.

Hence, $ABCD$ is a square.