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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
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