Question

Find the area of the square formed by (0, −1), (2, 1) (0, 3) and (−2, 1) taken in order are as vertices.

Answer 1

SOLUTION :  

Given :

Let A = (0, −1),B= (2, 1) C =(0, 3) D= (−2, 1)

Area of square ABCD = 2 × Area of ∆ABC

Area of ∆ = ½ |x1(y2-y3) +x2(y3 - y1) + x3(y3 - y1) |

Here, x1 =0, y1= -1, x2 = 2 ,y2= 1, x3= 0 ,y3 = 3

Area of ∆ ABC = ½|0(1 - 3) + 2(3-(-1))+ 0(-1-1)|

= ½ | 0× -2 + 2×4 + 0 × -2|

= ½| 0 + 8 +0|

= ½ × 8 = 4 sq.units

Area of ∆ ABC = 4 sq.units

Area of square ABCD = 2 × Area of ∆ABC

Area of square ABCD = 2 × 4 = 8 sq.units

Area of square ABCD = 8 sq.units

Hence, Area of square is 8 sq.units

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Let A ( 0, -1 ), B( 2,1 ),C ( 0, 3 ), and

D( -2 , 1 ) are the vertices of a Square.

we know that ,

AB = BC = CA = DA

Here ,

A( 0, -1 ) = ( x1, y1 ) ;

B ( 2 , 1 ) = ( x2 , y2 )

distance AB = √ ( x2 - x1 )² + ( y2 - y1 )²

AB = √ ( 2 - 0 )² + [ 1 - ( - 1 )]²

= √ 2² + 2²

AB = √8

Therefore ,

Area of Square = side²

= AB²

= ( √8 )²

= 8 square units.

I hope this helps you.

: )