Three vertices of a rectangle ABCD are A(3,1),B(-3,1)andC(-3,3). Plot these points on a graph paper and find the coordinates of the fourth vertex D . Also , find the area of rectangle ABCD.

Given: Three vertices of a rectangle ABCD are A$(3,1)$,B$(-3,1)$ and C$(-3,3)$.

To do: To find the area of rectangle ABCD.

Solution:

A(3,1), B(-3,1), C(-3,3)

Let say D = $(x , y)$

AD = BC

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

=>  $\sqrt{(x - 3)^2 +(y - 1)^2} =  4$

=> $x^2 + y^2 - 6x - 2y  = -6$

CD = AB

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

=>  $(x +3)² +(y - 3)² =  36$

=> $x² + y² + 6x -6 y  = 18$

=> $12x - 4y = 24$

=> $3x - y = 6$

BD=AC

=>  $(x +3)^2 +(y -1)^2 =  36 + 4$

We already got $(x +3)^2 +(y - 3)^2 =  36 =>(x +3)^2 = 36 - (y -3)^2$

=>$(y - 1)^2 -  (y - 3)^2 = 4$

=> $4y - 8 = 4$

=> $y =3$

$3x - y = 6$

=> $3x =9$

=> $x = 3$

$( 3, 3)$  is the coordinates of the fourth vertex.

Area = L$\times$B =AB$\times$BC

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

=  6 $\times$ 2 = 12 cm

Therefore the area of rectangle ABCD is 12cm