Question

Show that the points A(0, –2), B(3, 1), C(0, 4) and D(–3, 1) are vertices of a square. Also find its area.

Answer 1

Answer:

Area of the square ABCD is 18 square units

Step-by-step explanation:

Show that the points A(0, –2), B(3, 1), C(0, 4) and D(–3, 1) are vertices of a square. Also find its area.

Formula used:

$\text{The distance between two points }(x_1,y_1)\:and\:(x_2,y_2)\text{ is}$

$\boxed{\bf\:d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

Given points are

A(0, –2), B(3, 1), C(0, 4) and D(–3, 1)

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

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

$AB=\sqrt{9+9}$

$AB=3\sqrt{2}$

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

$BC=\sqrt{(3-0)^2+(1-4)^2}$

$BC=\sqrt{9+9}$

$BC=3\sqrt{2}$

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

$CD=\sqrt{(0+3)^2+(4-1)^2}$

$CD=\sqrt{9+9}$

$CD=3\sqrt{2}$

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

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

$AD=\sqrt{9+9}$

$AD=3\sqrt{2}$

$\implies\:AB=BC=CD=AD$

$\therefore\:\bf\text{All sides are equal}$

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

$AC=\sqrt{(0-0)^2+(-2-4)^2}$

$AC=\sqrt{0+36}$

$AC=6$

Now,

$AB^2+BC^2$

$=18+18$

'

$=36$

$=AC^2$

'

$\implies\:\angle{B}=90^{\circ}$

$\text{Hence ABCD is a square}$

$\text{Area of the square ABCD}$

$=AB^2=18\:\text{square units}$

Answer 2

the answer is given by my teacher.