Question

Prove that the given co-ordinates are vertices of a rectangle.

p(2,-2)
Q(8,4)
R(5,7)
S(-1,1)

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Answer 1

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 $.$

Answer 2

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