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