Prove that the given co-ordinates are vertices of a rectangle.
p(2,-2)
Q(8,4)
R(5,7)
S(-1,1)
Answers
Given :-
Here, A quadrilateral PQRS is given
The coordinates of point P ( 2 , -2 ) , Q( 8,4 ),
R( 5,7 ) and S( -1 , 1 ) are given
Solution :-
We have to prove that quadrilateral PQRS is a rectangle.
Therefore ,
By using distance formula
= √ ( x2 - x1 )^2 + ( y2 - y1 )^2
Compare the all four vertices of a quadrilateral with ( x1 , y1 ) , ( x2 , y2 )
Therefore,
PQ = √ ( 8 - 2 )^2 + ( 4 - ( -2)^2
PQ = √ ( 8 - 2 )^2 + ( 4 + 2 )^2
PQ = √ ( 6 )^2 + ( 6 )^2
PQ = √ 36 + 36
PQ = √72
PQ = 6√2
QR = √ ( 5 - 8 )^2 + ( 7 - 4 )^2
QR = √ ( -3 )^2 + ( 3 )^2
QR = √ 9 + 9
QR = √18
QR = 3√2
RS = √( -1 - 5 )^2 + ( 1 - 7 )^2
RS = √( -6)^2 + ( 6 )^2
RS = √36 + 36
RS = √72
RS = 6√2
SP = √( 2- (-1))^2 + ( -2 - 1 )^2
SP = √( 2 + 1 )^2 + ( -2 - 1 )^2
SP = √( 3 )^2 + (-3)^2
SP = √9+ 9
SP = √18
SP = 3√2
Now,
PR = √( 5 - 2 )^2 + ( 7 - ( -2 )^2
PR = √(5 -2 )^2 + ( 7 + 2 )^2
PR = √( 3)^2 + ( 9)^2
PR = √9 + 81
PR = √90
PR = 3√10
QS = √(-1 - 8 )^2+ ( 1 - 4 )^2
QS = √( -9)^2 + ( -3 )^2
QS = √81 + 9
QS = √90
QS =3√10
Here, You can observe that ,
PQ = RS, QR = SP
[ It means opposite sides are equal ]
Here, PR = QS
[ Diagonals are equal ]
Hence, PQRS is a rectangle 
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